Sign homomorphism (from the symmetric group)

The sign [47t homomorphism] is given by sending a permutation $\sigma$ in the [-497] $S_n$ to $0$ if we can make $\sigma$ by multiplying together an even number of [4cn transpositions], and to $1$ otherwise.

%%%knows-requisite([modular_arithmetic]):
Equivalently, it is given by sending $\sigma$ to the number of transpositions making it up, [modular_arithmetic modulo] $2$.
%%%

The sign homomorphism [4hh is well-defined].

%%%knows-requisite([quotient_group]):
The [-alternating_group] is obtained by taking the [quotient_group quotient] of the symmetric group by the sign homomorphism.
%%%


Alternating group is generated by its three-cycles

The [-4hf] $A_n$ is generated by its $3$-[49f cycles].
That is, every element of $A_n$ can be made by multiplying together $3$-cycles only.

# Proof

The product of two [4cn transpositions] is a product of $3$-cycles:

- $(ij)(kl) = (ijk)(jkl)$
- $(ij)(jk) = (ijk)$
- $(ij)(ij) = e$.

Therefore any permutation which is a product of evenly-many transpositions (that is, all of $A_n$) is a product of $3$-cycles, because we can group up successive pairs of transpositions.

Conversely, every $3$-cycle is in $A_n$ because $(ijk) = (ij)(jk)$.


Group coset

Given a subgroup $H$ of [-3gd] $G$, the *left cosets* of $H$ in $G$ are sets of the form $\{ gh : h \in H \}$, for some $g \in G$.
This is written $gH$ as a shorthand.

Similarly, the *right cosets* are the sets of the form $Hg = \{ hg: h \in H \}$.

# Examples
%%%knows-requisite([497]):
## Symmetric group

In $S_3$, the [-497] on three elements, we can list the elements as $\{ e, (123), (132), (12), (13), (23) \}$, using [49f cycle notation].
Define $A_3$ (which happens to have a name: the [-4hf]) to be the subgroup with elements $\{ e, (123), (132) \}$.

Then the coset $(12) A_3$ has elements $\{ (12), (12)(123), (12)(132) \}$, which is simplified to $\{ (12), (23), (13) \}$.

The coset $(123)A_3$ is simply $A_3$, because $A_3$ is a subgroup so is closed under the group operation. $(123)$ is already in $A_3$.
%%%

[todo: more examples, with different requirements]

# Properties

- The left cosets of $H$ in $G$ [set_partition partition] $G$. ([4j5 Proof.])
- For any pair of left cosets of $H$, there is a [499 bijection] between them; that is, all the cosets are all the same size. ([4j8 Proof.])

# Why are we interested in cosets?

Under certain conditions (namely that the subgroup $H$ must be [4h6 normal]), we may define the [-quotient_group], a very important concept; see the page on [4j5 "left cosets partition the parent group"] for a glance at why this is useful.
[todo: there must be a less clumsy way to do it]

Additionally, there is a key theorem whose usual proof considers cosets ([lagrange_theorem_on_subgroup_size Lagrange's theorem]) which strongly restricts the possible sizes of subgroups of $G$, and which itself is enough to classify all the groups of [3gg order] $p$ for $p$ [prime_number prime].
Lagrange's theorem also has very common applications in [-number_theory], in the form of the [fermat_euler_theorem Fermat-Euler theorem].


Left cosets partition the parent group

Given a [-3gd] $G$ and a subgroup $H$, the [4j4 left cosets] of $H$ in $G$ [set_partition partition] $G$, in the sense that every element of $g$ is in precisely one coset.

# Proof
Firstly, every element is in a coset: since $g \in gH$ for any $g$.
So we must show that no element is in more than one coset.

Suppose $c$ is in both $aH$ and $bH$.
Then we claim that $aH = cH = bH$, so in fact the two cosets $aH$ and $bH$ were the same.
Indeed, $c \in aH$, so there is $k \in H$ such that $c = ak$.
Therefore $cH = \{ ch : h \in H \} = \{ akh : h \in H \}$.

Exercise: $\{ akh : h \in H \} = \{ ar : r \in H \}$.
%%hidden(Show solution):
Suppose $akh$ is in the left-hand side.
Then it is in the right-hand side immediately: letting $r=kh$.

Conversely, suppose $ar$ is in the right-hand side.
Then we may write $r = k k^{-1} r$, so $a k k^{-1} r$ is in the right-hand side; but then $k^{-1} r$ is in $H$ so this is exactly an object which lies in the left-hand side.
%%

But that is just $aH$.

By repeating the reasoning with $a$ and $b$ interchanged, we have $cH = bH$; this completes the proof.

# Why is this interesting?

The fact that the left cosets partition the group means that we can, in some sense, "compress" the group $G$ with respect to $H$.
If we are only interested in $G$ "up to" $H$, we can deal with the partition rather than the individual elements, throwing away the information we're not interested in.

This concept is most importantly used in defining the [-4tq].
To do this, the subgroup must be [4h6 normal] ([4h9 proof]).
In this case, the collection of cosets itself inherits a group structure from the parent group $G$, and the structure of the quotient group can often tell us a lot about the parent group.


Church-Turing thesis: Evidence for the Church-Turing thesis

As the Church-Turing thesis is not a [ proper mathematical sentence] we cannot prove it. However, we can collect [-3s] to increase our confidence in its correctness.

#The inductive argument
Every computational model we have seen so far is [ reducible] to Turing's model.

Indeed, the thesis was originally formulated independently by Church and Turing in reference to two different computational models ([Turing_machines] and [Lambda_Calculus] respectively). When they were shown to be [equivilance_relation equivalent] it was massive evidence in favor of both of them.

A non-exhaustive list of models which can be shown to be reducible to Turing machines are:

* [Lambda_calculus]
* [Quantum_computation]
* [Nondeterministic_Turing_machines Non-deterministic_Turing_machines]
* [Register_machines]
* The set of [-recursive_functions]

#Lack of counterexamples
Perhaps the strongest argument we have for the CT thesis is that there is not a widely accepted candidate to a counterexample of the thesis. This is unlike the [-4ct], where quantum computation stands as a likely counterexample.

One may wonder whether the computational models which use a source of [ randomness] (such as quantum computation or [-probabilistic_Turing_machines]) are a proper counterexample to the thesis: after all, Turing machines are fully [-deterministic], so they cannot simulate randomness.

To properly explain this issue we have to recall what it means for a quantum computer or a probabilistic Turing machine to compute something: we say that such a device computes a function $f$ if for every input $x$ the probability of the device outputting $f(x)$ is greater or equal to some arbitrary constant greater than $1/2$. Thus we can compute $f$ in a classical machine, for there exists always the possibility of simulating every possible outcome of the randomness to deterministically compute the probability distribution of the output, and output as a result the possible outcome with greater probability.

Thus, randomness is reducible to Turing machines, and the CT thesis holds.


Left cosets are all in bijection

Let $H$ be a subgroup of $G$.
Then for any two [4j4 left cosets] of $H$ in $G$, there is a [-499] between the two cosets.

# Proof

Let $aH, bH$ be two cosets.
Define the function $f: aH \to bH$ by $x \mapsto b a^{-1} x$.

This has the correct [3lg codomain]: if $x \in aH$ (so $x = ah$, say), then $ba^{-1} a x = bx$ so $f(x) \in bH$.

The function is [4b7 injective]: if $b a^{-1} x = b a^{-1} y$ then (pre-multiplying both sides by $a b^{-1}$) we obtain $x = y$.

The function is [4bg surjective]: given $b h \in b H $, we want to find $x \in aH$ such that $f(x) = bh$.
Let $x = a h$ to obtain $f(x) = b a^{-1} a h = b h$, as required.


The alternating groups on more than four letters are simple

Let $n > 4$ be a [-45h]. Then the [-4hf] $A_n$ on $n$ elements is [simple_group simple].

# Proof

We go by [mathematical_induction induction] on $n$.
The base case of $n=5$ [4jf we treat separately].

For the inductive step: let $n \geq 6$, and suppose $H$ is a nontrivial [-4h6] of $A_n$.
Ultimately we aim to show that $H$ contains the $3$-[49f cycle] $(123)$; since $H$ [subgroup_normal_iff_union_of_conjugacy_classes is a union of conjugacy classes], and since the $3$-cycles [splitting_conjugacy_classes_in_alternating_group form a conjugacy class] in $A_n$, this means every $3$-cycle is in $H$ and [4hs hence] $H = A_n$.

## Lemma: $H$ contains a member of $A_{n-1}$
To start, we will show that at least $H$ contains *something* from $A_{n-1}$ (which we view as all the elements of $A_n$ which don't change the letter $n$). %%note:Recall that $A_n$ [3t9 acts] naturally on the set of "letters" $\{1,2,\dots, n \}$ by permutation.%%
(This approach has to be useful for an induction proof, because we need some way of introducing the simplicity of $A_{n-1}$.)
That is, we will show that there is some $\sigma \in H$ with $\sigma \not = e$, such that $\sigma(n) = n$.

Let $\sigma \in H$, where $\sigma$ is not the identity.
$\sigma$ certainly sends $n$ somewhere; say $\sigma(n) = i$, where $i \not = n$ (since if it is, we're done immediately).

Then if we can find some $\sigma' \in H$, not equal to $\sigma$, such that $\sigma'(n) = i$, we are done: $\sigma^{-1} \sigma'(n) = n$.

$\sigma$ must move something other than $i$ (that is, there must be $j \not = i$ such that $\sigma(j) \not = j$), because if it did not, it would be the transposition $(n i)$, which is not in $A_n$ because it is an odd number of transpositions.
Hence $\sigma(j) \not = j$ and $j \not = i$; also $j \not = n$ because if it were, [todo: this]

Now, since $n \geq 6$, we can pick $x, y$ distinct from $n, i, j, \sigma(j)$.
Then set $\sigma' = (jxy) \sigma (jxy)^{-1}$, which must lie in $H$ because $H$ is closed under [4gk conjugation].

Then $\sigma'(n) = i$; and $\sigma' \not = \sigma$, because $\sigma'(j) = \sigma(y)$ which is not equal to $\sigma(j)$ (since $y \not = j$).
Hence $\sigma'$ and $\sigma$ have different effects on $j$ so they are not equal.

## Lemma: $H$ contains all of $A_{n-1}$
Now that we have shown $H$ contains some member of $A_{n-1}$.
But $H \cap A_{n-1}$ is normal in $A_n$, because $H$ is normal in $A_n$ and it is [4h6 _intersect _subgroup _is _normal the intersection of a subgroup with a normal subgroup].
Therefore by the inductive hypothesis, $H \cap A_{n-1}$ is either the trivial group or is $A_{n-1}$ itself.

But $H \cap A_{n-1}$ is certainly not trivial, because our previous lemma gave us a non-identity element in it; so $H$ must actually contain $A_{n-1}$.

## Conclusion

Finally, $H$ contains $A_{n-1}$ so it contains $(123)$ in particular; so we are done by the initial discussion.

# Behaviour for $n \leq 4$
- $A_1$ is the trivial group so is vacuously not simple.
- $A_2$ is also the trivial group.
- $A_3$ is isomorphic to the [-47y] $C_3$ on three generators, so it is simple: it has no nontrivial proper subgroups, let alone normal ones.
- $A_4$ has the following normal subgroup (the [klein_four_group Klein four-group]): $\{ e, (12)(34), (13)(24), (14)(23) \}$. Therefore $A_4$ is not simple.


Simple group

A simple group is a [-3gd] with no [4h6 normal subgroups].
[todo: everything]


The alternating group on five elements is simple

The [-4hf] $A_5$ on five elements is [4jc simple].

# Proof

Recall that $A_5$ has [3gg order] $60$, so [4jn Lagrange's theorem] states that any subgroup of $A_5$ has order dividing $60$.

Suppose $H$ is a normal subgroup of $A_5$, which is not the trivial subgroup $\{ e \}$.
If $H$ has order divisible by $3$, then by [4l6 Cauchy's theorem] there is a $3$-[49f cycle] in $H$ (because the $3$-cycles are the only elements with order $3$ in $A_5$).
Because $H$ [4jw is a union of conjugacy classes], and because the $3$-cycles [4kv form a conjugacy class in $A_n$ for $n > 4$], $H$ would therefore contain *every* $3$-cycle; but then [4hs it would be the entire alternating group].

If instead $H$ has order divisible by $2$, then there is a double transposition such as $(12)(34)$ in $H$, since these are the only elements of order $2$ in $A_5$.
But then $H$ contains the entire conjugacy class so it contains every double transposition; in particular, it contains $(12)(34)$ and $(15)(34)$, so it contains $(15)(34)(12)(34) = (125)$.
Hence as before $H$ contains every $3$-cycle so is the entire alternating group.

So $H$ must have order exactly $5$, by [4jn Lagrange's theorem]; so it contains an element of order $5$ since [4jh prime order groups are cyclic].

The only such elements of $A_n$ are $5$-cycles; but the conjugacy class of a $5$-cycle is of size $12$, which is too big to fit in $H$ which has size $5$.


Prime order groups are cyclic

Let $G$ be a [-3gd] whose [3gg order] is equal to $p$, a [-4mf].
Then $G$ is [49x isomorphic] to the [-47y] $C_p$ of order $p$.

# Proof
Pick any non-identity element $g$ of the group.

By [4jn Lagrange's theorem], the subgroup generated by $g$ has size $1$ or $p$ (since $p$ was prime).
But it can't be $1$ because the only subgroup of size $1$ is the trivial subgroup.

Hence the subgroup must be the entire group.


Lagrange theorem on subgroup size

Lagrange's Theorem states that if $G$ is a finite [-3gd] and $H$ a subgroup, then the [3gg order] $|H|$ of $H$ divides the order $|G|$ of $G$.
It generalises to infinite groups: the statement then becomes that the [4j4 left cosets] form a [set_partition partition], and for any pair of cosets, there is a [499 bijection] between them.

# Proof

In full generality, the cosets [4j5 form a partition] and [4j8 are all in bijection].

To specialise this to the finite case, we have divided the $|G|$ elements of $G$ into buckets of size $|H|$ (namely, the cosets), so $|G|/|H|$ must in particular be an integer.


The alternating group on five elements is simple: Simpler proof

If there is a non-trivial [-4h6] $H$ of the [-4hf] $A_5$ on five elements, then [4jw it is a union of conjugacy classes].
Additionally, by [4jn Lagrange's theorem], the [3gg order] of a subgroup must divide the order of the group, so the total size of $H$ must divide $60$.

We can list the [alternating_group_five_conjugacy_classes conjugacy classes of $A_5$]; they are of size $1, 20, 15, 12, 12$ respectively.

By a brute-force check, no sum of these containing $1$ can possibly divide $60$ (which is the size of $A_5$) unless it is $1$ or $60$.

# The brute-force check

We first list the [number_theory_divisor divisors] of $60$: they are $$1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$$

Since the subgroup $H$ must contain the identity, it must contain the conjugacy class of size $1$.
If it contains any other conjugacy class (which, as it is a non-trivial subgroup by assumption, it must), then the total size must be at least $13$ (since the smallest other class is of size $12$); so it is allowed to be one of $15$, $20$, $30$, or $60$.
Since $H$ is also assumed to be a *proper* subgroup, it cannot be $A_5$ itself, so in fact $60$ is also banned.

## The class of size $20$
If $H$ then contains the conjugacy class of size $20$, then $H$ can only be of size $30$ because we have already included $21$ elements.
But there is no way to add just $9$ elements using conjugacy classes of size bigger than or equal to $12$.

So $H$ cannot contain the class of size $20$.

## The class of size $15$

In this case, $H$ is allowed to be of size $20$ or $30$, and we have already found $16$ elements of it. So there are either $4$ or $14$ elements left to find; but we are only allowed to add classes of size exactly $12$, so this can't be done either.

## The classes of size $12$

What remains is two classes of size $12$, from which we can make $1+12 = 13$ or $1+12+12 = 25$.
Neither of these divides $60$, so these are not legal options either.

This exhausts the search, and completes the proof.


Subgroup is normal if and only if it is a union of conjugacy classes

Let $H$ be a subgroup of the [-3gd] $G$.
Then $H$ is [4h6 normal] in $G$ if and only if it can be expressed as a [set_union union] of [4bj conjugacy classes].

# Proof

$H$ is normal in $G$ if and only if $gHg^{-1} = H$ for all $g \in G$; equivalently, if and only if $ghg^{-1} \in H$ for all $h \in H$ and $g \in G$.

But if we fix $h \in H$, then the statement that $ghg^{-1} \in H$ for all $g \in G$ is equivalent to insisting that the conjugacy class of $h$ is contained in $H$.
Therefore $H$ is normal in $G$ if and only if, for all $h \in H$, the conjugacy class of $h$ lies in $H$.

If $H$ is normal, then it is clearly a union of conjugacy classes (namely $\cup_{h \in H} C_h$, where $C_h$ is the conjugacy class of $h$).

Conversely, if $H$ is not normal, then there is some $h \in H$ such that the conjugacy class of $h$ is not wholly in $H$; so $H$ is not a union of conjugacy classes because it contains $h$ but not the entire conjugacy class of $h$.
(Here we have used that the [-conjugacy_classes_partition_the_group].)

# Interpretation

A normal subgroup is one which is fixed under conjugation; the most natural (and, indeed, the smallest) objects which are fixed under conjugation are conjugacy classes; so this criterion tells us that to obtain a *subgroup* which is fixed under conjugation, it is necessary and sufficient to assemble these objects (the conjugacy classes), which are themselves the smallest objects which are fixed under conjugation, into a group.


Conjugacy classes of the alternating group on five elements

This page lists the [4bj conjugacy classes] of the [-4hf] $A_5$ on five elements.
See a [4l0 different lens] for a derivation of this result using less theory.

$A_5$ has size $5!/2 = 60$, where the exclamation mark denotes the [-factorial] function.
We will assume access to [4bk the conjugacy class table of $S_5$] the [-497] on five elements; $A_5$ is a [quotient_group quotient] of $S_5$ by the [4hk sign homomorphism].

We have that a conjugacy class splits if and only if its [49f cycle type] is all odd, all distinct. ([4kv Proof.])
This makes the classification of conjugacy classes very easy.

# The table

We must remove all the lines of [4bk $S_5$'s table] which correspond to odd permutations (that is, those which are the product of odd-many [4cn transpositions]). Indeed, those lines are classes which are not even in $A_5$.

We are left with cycle types $(5)$, $(3, 1, 1)$, $(2, 2, 1)$, $(1,1,1,1,1)$.
Only the $(5)$ cycle type can split into two, by the splitting condition.
It splits into the class containing $(12345)$ and the class which is $(12345)$ conjugated by odd permutations in $S_5$.
A representative for that latter class is $(12)(12345)(12)^{-1} = (21345)$.

$$\begin{array}{|c|c|c|c|}
\hline
\text{Representative}& \text{Size of class} & \text{Cycle type} & \text{Order of element} \\ \hline
(12345) & 12 & 5 & 5 \\ \hline
(21345) & 12 & 5 & 5 \\ \hline
(123) & 20 & 3,1,1 & 3 \\ \hline
(12)(34) & 15 & 2,2,1 & 2 \\ \hline
e & 1 & 1,1,1,1,1 & 1 \\ \hline
\end{array}$$


Splitting conjugacy classes in alternating group

Recall that in the [-497] $S_n$, the notion of "[-4bj]" coincides with that of "has the same [4cg cycle type]" ([4bh proof]).
It turns out to be the case that when we descend to the [-4hf] $A_n$, the conjugacy classes nearly all remain the same; the exception is those classes in $S_n$ which have cycle type all odd and all distinct.
Those classes split into exactly two classes in $A_n$.

# Proof

## Conjugate implies same cycle type

This direction of the proof is identical to [4bh the same direction] in the proof of the corresponding result on symmetric groups.

## Converse: splitting condition

Recall before we start that an even-length cycle can only be written as the product of an *odd* number of transpositions, and vice versa.

The question is: is every $\tau \in A_n$ with the same cycle type as $\sigma \in A_n$ conjugate (in $A_n)$ to $\sigma$?
If so, then the conjugacy class in $A_n$ of $\sigma$ is just the same as that of $\sigma$ in $S_n$; if not, then the conjugacy class in $S_n$ must break into two pieces in $A_n$, namely $\{ \rho \sigma \rho^{-1} : \rho \ \text{even} \}$ and $\{ \rho \sigma \rho^{-1} : \rho \ \text{odd} \}$.
(Certainly these are conjugacy classes: they only contain even permutations so they are subsets of $A_n$, while everything in either class is conjugate in $A_n$ because the definition only depends on the [even_odd_parity parity] of $\rho$.)

Let $\sigma = c_1 \dots c_k$, and $\tau = c_1' \dots c_k'$ of the same cycle type: $c_i = (a_{i1} \dots a_{i r_i})$, $c_i' = (b_{i1} \dots b_{i r_i})$.

Define $\rho$ to be the permutation which takes $a_{ij}$ to $b_{ij}$, but otherwise does not move any other letter.
Then $\rho \sigma \rho^{-1} = \tau$, so if $\rho$ lies in $A_n$ then we are done: $\sigma$ and $\tau$ are conjugate in $A_n$.

That is, we may assume without loss of generality that $\rho$ is odd.

### If any of the cycles is even in length

Suppose without loss of generality that $r_1$, the length of the first cycle, is even.
Then we can rewrite $c_1' = (b_{11} \dots b_{1r_1})$ as $(b_{12} b_{13} \dots b_{1 r_1} b_{11})$, which is the same cycle expressed slightly differently.

Now $c_1'$ is even in length, so it is an odd permutation (being a product of an odd number of transpositions, $(b_{1 r_1} b_{11}) (b_{1 (r_1-1)} b_{11}) \dots (b_{13} b_{11})(b_{12} b_{11})$).
Hence $\rho c_1'$ is an even permutation.

But conjugating $\tau$ by $\rho c_1'$ yields $\sigma$:
$$\sigma = \rho \tau \rho^{-1} = \rho c_1' (c_1'^{-1} \tau c_1') c_1'^{-1} \rho^{-1}$$
which is the result of conjugating $c_1'^{-1} \tau c_1' = \tau$ by $\rho c_1'$.

(It is the case that $c_1'^{-1} \tau c_1' = \tau$, because $c_1'$ commutes with all of $\tau$ except with the first cycle $c_1'$, so the expression is $c_1'^{-1} c_1' c_1' c_2' \dots c_k'$, where we have pulled the final $c_1'$ through to the beginning of the expression $\tau = c_1' c_2' \dots c_k'$.)

%%hidden(Example):
Suppose $\sigma = (12)(3456), \tau = (23)(1467)$.

We have that $\tau$ is taken to $\sigma$ by conjugating with $\rho = (67)(56)(31)(23)(12)$, which is an odd permutation so isn't in $A_n$.
But we can rewrite $\tau = (32)(1467)$, and the new permutation we'll conjugate with is $\rho c_1' = (67)(56)(31)(23)(12)(32)$, where we have appended $c_1' = (23) = (32)$.
It is the case that $\rho$ is an even permutation and hence is in $A_n$, because it is the result of multiplying the odd permutation $\rho$ by the odd permutation $c_1'$.

Now the conjugation is the composition of two conjugations: first by $(32)$, to yield $(32)\tau(32)^{-1} = (23)(1467)$ (which is $\tau$ still!), and then by $\rho$.
But we constructed $\rho$ so as to take $\tau$ to $\sigma$ on conjugation, so this works just as we needed.
%%

## If all the cycles are of odd length, but some length is repeated
Without loss of generality, suppose $r_1 = r_2$ (and label them both $r$), so the first two cycles are of the same length: say $\sigma$'s version is $(a_1 a_2 \dots a_r)(c_1 c_2 \dots c_r)$, and $\tau$'s version is $(b_1 b_2 \dots b_r)(d_1 d_2 \dots d_r)$.

Then define $\rho' = \rho (b_1 d_1)(b_2 d_2) \dots (b_r d_r)$.
Since $r$ is odd and $\rho$ is an odd permutation, $\rho'$ is an even permutation.

Now conjugating by $\rho'$ is the same as first conjugating by $(b_1 d_1)(b_2 d_2) \dots (b_r d_r)$ and then by $\rho$.

But conjugating by $(b_1 d_1)(b_2 d_2) \dots (b_r d_r)$ takes $\tau$ to $\tau$, because it has the effect of replacing all the $b_i$ by $d_i$ and all the $d_i$ by $b_i$, so it simply swaps the two cycles round.

Hence the conjugation of $\tau$ by $\rho'$ yields $\sigma$, and $\rho'$ is in $A_n$.

%%hidden(Example):
Suppose $\sigma = (123)(456), \tau = (154)(237)$.

Then conjugation of $\tau$ by $\rho = (67)(35)(42)(34)(25)$, an odd permutation, yields $\sigma$.

Now, if we first perform the conjugation $(12)(53)(47)$, we take $\tau$ to itself, and then performing $\rho$ yields $\sigma$.
The combination of $\rho$ and $(12)(53)(47)$ is an even permutation, so it does line in $A_n$.
%%

## If all the cycles are of odd length, and they are all distinct

In this case, we are required to check that the conjugacy class *does* split.
Remember, we started out by supposing $\sigma$ and $\tau$ have the same cycle type, and they are conjugate in $S_n$ by an odd permutation (so they are not obviously conjugate in $A_n$); we need to show that indeed they are *not* conjugate in $A_n$.

Indeed, the only ways to rewrite $\tau$ into $\sigma$ (that is, by conjugation) involve taking each individual cycle and conjugating it into the corresponding cycle in $\sigma$. There is no choice about which $\tau$-cycle we take to which $\sigma$-cycle, because all the cycle lengths are distinct.
But the permutation $\rho$ which takes the $\tau$-cycles to the $\sigma$-cycles is odd, so is not in $A_n$.

Moreover, since each cycle is odd, we can't get past the problem by just cycling round the cycle (for instance, by taking the cycle $(123)$ to the cycle $(231)$), because that involves conjugating by the cycle itself: an even permutation, since the cycle length is odd.
Composing with the even permutation can't take $\rho$ from being odd to being even.

Therefore $\tau$ and $\sigma$ genuinely are not conjugate in $A_n$, so the conjugacy class splits.

%%hidden(Example):
Suppose $\sigma = (12345)(678), \tau = (12345)(687)$ in $A_8$.

Then conjugation of $\tau$ by $\rho = (87)$, an odd permutation, yields $\sigma$.
Can we do this with an *even* permutation instead?

Conjugating $\tau$ by anything at all must keep the cycle type the same, so the thing we conjugate by must take $(12345)$ to $(12345)$ and $(687)$ to $(678)$.

The only ways of $(12345)$ to $(12345)$ are by conjugating by some power of $(12345)$ itself; that is even.
The only ways of taking $(687)$ to $(678)$ are by conjugating by $(87)$, or by $(87)$ and then some power of $(678)$; all of these are odd.

Therefore the only possible ways of taking $\tau$ to $\sigma$ involve conjugating by an odd permutation $(678)^m(87)$, possibly alongside some powers of an even permutation $(12345)$; therefore to get from $\tau$ to $\sigma$ requires an odd permutation, so they are in fact not conjugate in $A_8$.
%%

# Example

In $A_7$, the cycle types are $(7)$, $(5, 1, 1)$, $(4,2,1)$, $(3,2,2)$, $(3,3,1)$, $(3,1,1,1,1)$, $(1,1,1,1,1,1,1)$, and $(2,2,1,1,1)$.
The only class which splits is the $7$-cycles, of cycle type $(7)$; it splits into a pair of half-sized classes with representatives $(1234567)$ and $(12)(1234567)(12)^{-1} = (2134567)$.


Decision problem

[summary:  A decision problem is a question of the form "does [-bitstring] $w$ have [-property] $p$, yes or no?"]

A **decision problem** is a subset $D$ of a set of instances $A$, where generally $A$ is the set of finite [-bitstrings] $\{0,1\}^*$.

In plain English, a decision problem is composed by a number of members of a collection, which generally share a common property.

In plainer English, a decision problem is a question of the form "does bitstring $w$ have property $p$, yes or no?"

Every question that fits the set form (is this string in the set?) can be re-expressed as "does this bitstring have the property of being in the set?" Everything that fits the property form can be re-expressed in set form as well, because a property implicitly specifies a set of things that have the property. Thus, the two forms are equivalent.

It is called a decision problem because we are trying to *decide* whether a given bitstring belongs to the set or not.

A **solution** of the problem would consist in a procedure which given an instance $w$ of $A$ decides correctly in finite time whether $w$ belongs to $D$ or not. That is, a solution consists in a way of identifying the common trait shared by the members of $D$ which distinguish them from the rest of instances in $A$.

We say that a problem is [-decidable] if there exists a solution which works for every instance. Trivially, finite decision problems are decidable, since we can have a look-up list with every element which we would consult every time we are given an instance to classify. If the instance corresponds to an element in the list, we accept it, and otherwise reject it.

We say that a decision problem is [-semidecidable] if there is a procedure which, in finite time, identifies members of $D$, but fails to [-halt] and reject instances which do not belong to $D$.

Decision problems can be classified according to their difficulty of solving in [ complexity classes], and are a central object of study in [-49w].

#Examples
##Graph connectedness
Suppose we encode every possible finite graph, up to isomorphism, in the collection of finite bitstrings. Then we could define:
$$
CONNECTED = \{s\in\{0,1\}^*:\text{$s$ represents a connected graph}\}
$$
Which would be a [graph_connectedness decidable decision problem].

##Tautology identification
Let $TAUTOLOGY$ be the collection of formulas of [-first_order_logic] which are true for every [mathematical_interpretation interpretation]. Then [ $TAUTOLOGY$ is a semidecidable decision problem], as if a formula is [-valid] we can search for a proof, but otherwise, we do not have an effective procedure to decide that a formula is not valid.

##Primality
Let:
$$
PRIMES = \{ x\in \mathbb{N}:\text{$x$ is prime}\}
$$
Then $PRIMES$ is a decidable decision problem.

We can alternately define
$$
PRIMES = \{s\in\{0,1\}^*:\text{$s$ represent a prime number in base $2$}\}
$$
This illustrates that we can specify a particular decision problem in different ways.


Conjugacy classes of the alternating group on five elements: Simpler proof

The [-4hf] $A_5$ on five elements has [4bj conjugacy classes] very similar to those of the [-497] $S_5$, but where one of the classes has split in two.

Note that $A_5$ has $60$ elements, since it is precisely half of the elements of $S_5$ which has $5! = 120$ elements (where the exclamation mark is the [-factorial] function).

# Table

$$\begin{array}{|c|c|c|c|}
\hline
\text{Representative}& \text{Size of class} & \text{Cycle type} & \text{Order of element} \\ \hline
(12345) & 12 & 5 & 5 \\ \hline
(21345) & 12 & 5 & 5 \\ \hline
(123) & 20 & 3,1,1 & 3 \\ \hline
(12)(34) & 15 & 2,2,1 & 2 \\ \hline
e & 1 & 1,1,1,1,1 & 1 \\ \hline
\end{array}$$

# Working

Firstly, the identity is in a class of its own, because $\tau e \tau^{-1} = \tau \tau^{-1} = e$ for every $\tau$.

Now, by the same reasoning as in [4bh the proof] that conjugate elements must have the same cycle type in $S_n$, that result also holds in $A_n$.

Hence we just need to see whether any of the cycle types comprise more than one conjugacy class.

Recall that the available cycle types are $(5)$, $(3,1,1)$, $(2,2,1)$, $(1,1,1,1,1)$ (the last of which is the identity and we have already considered it).

## Double-transpositions

All the double-transpositions are conjugate (so the $(2,2,1)$ cycle type does not split):

- $(ab)(cd)$ is conjugate to $(ab)(ce)$ if we conjugate by $(ab)(de)$; symmetrically this covers all the cases where one of the two transpositions remains the same.
- $(ab)(cd)$ is conjugate to $(ac)(bd)$ by $(cba)$; this covers the case that $e$ is not introduced.
- $(ab)(cd)$ is conjugate to $(ac)(be)$ by $(bc)(de)$; this covers the remaining cases that $e$ is introduced and neither of the two transpositions remains the same.

## Three-cycles

All the three-cycles are conjugate (so the $(3,1,1)$ cycle type does not split):

- $(abc)$ is conjugate to $(acb)$ by $(bc)(de)$, so three-cycles are conjugate to their permutations.
- $(abc)$ is conjugate to $(abd)$ by $(cde)$; this covers the case of introducing a single new element to the cycle.
- $(abc)$ is conjugate to $(ade)$ by $(bd)(ce)$; this covers the case of introducing two new elements to the cycle.

## Five-cycles

This class does split: I claim that $(12345)$ and $(21345)$ are not conjugate.
(Once we have this, then the class must split into two chunks, since $\{ \rho (12345) \rho^{-1}: \rho \ \text{even} \}$ is closed under conjugation in $A_5$, and $\{ \rho (12345) \rho^{-1}: \rho \ \text{odd} \}$ is closed under conjugation in $A_5$.
The first is the conjugacy class of $(12345)$ in $A_5$; the second is the conjugacy class of $(21345) = (12)(12345)(12)^{-1}$.
The only question here was whether they were separate conjugacy classes or whether their union was the conjugacy class.)

Recall that $\tau (12345) \tau^{-1} = (\tau(1), \tau(2), \tau(3), \tau(4), \tau(5))$, so we would need a permutation $\tau$ such that $\tau$ sends $1$ to $2$, $2$ to $1$, $3$ to $3$, $4$ to $4$, and $5$ to $5$.
The only such permutation is $(12)$, the transposition, but that is not actually a member of $A_5$.

Hence in fact $(12345)$ and $(21345)$ are not conjugate.


Poset: Exercises

Try these exercises to test your poset knowledge.

# Corporate Ladder

Imagine a company with five distinct roles: CEO, marketing manager, marketer, IT manager, and IT worker. At this company, if the CEO gives an order, everyone else must follow it. If an order comes from an IT manager, only IT workers are obligated to follow it. Similarly, if an order comes from a marketing manager, only marketers are obligated to follow it. Nobody is obligated to follow orders from marketers or IT workers.

Do the workers at this company form a poset under the "obligated to follow orders from" relation?

%%%hidden(Show solution):
While not technically a poset due to its lack of reflexivity, it is pretty close. It is actually a strict ordering, whose underlying partial order could be obtained by making the reasonable assumption that each worker will follow her own orders.
%%%

# Bag Inclusion

We can define a notion of [power_set power sets] for [3jk bags] as follows. Let $X$ be a set, then we use $\mathcal{M}(X)$ to denote the set of all bags containing elements of $X$. Let $A \in \mathcal{M}(X)$. The multiplicity function of $A$, $1_A : X \rightarrow \mathbb N$ maps each element of $X$ to the number of times that element occurs in $A$. We can use multiplicity functions to define a inclusion relation $\subseteq$ for bags. For $A, B \in \mathcal M(X)$, we write $A \subseteq B$ whenever for all $x \in X$, $1_A(x) \leq 1_B(x)$.

Does $\mathcal{M}(X)$ form a poset under the bag inclusion relation $\subseteq$? If so, prove it. Otherwise, show that it does not satisfy one of the three poset properties.

# Duality

Give the dual of the following proposition.

For all posets $P$ and all $p, q \in P$, $q \prec p$ implies that $\{ r \in P~|~r \leq p \}$ is a superset of $\{ r \in P~|~r \leq q\}$.

%%hidden(Show solution):
For all posets $P$ and all $p, q \in P$, $q \succ p$ implies that $\{ r \in P~|~r \geq p \}$ is a superset of $\{ r \in P~|~r \geq q\}$ (where $q \succ p$ means $p \prec q$).
%%

# Hasse diagrams

Let $X = \{ x, y, z \}$. Draw a Hasse diagram for the poset $\langle \mathcal P(X), \subseteq \rangle$ of the power set of $X$ ordered by inclusion.

%%hidden(Show solution):
![A Hasse diagram of the power set of X, ordered by inclusion](http://i.imgur.com/WG3OLFc.png)

%%%comment:
dot source:

digraph G {
  node [width = 0.1, height = 0.1]
  edge [arrowhead = "none"]
  e [label = "{}"]
  x [label = "{x}"]
  y [label = "{y}"]
  z [label = "{z}"]
  xy [label = "{x,y}"]
  xz [label = "{x,z}"]
  yz [label = "{y,z}"]
  xyz [label = "{x,y,z}"]

  rankdir = BT;
  e -> x
  e -> y
  e -> z
  x -> xy
  x -> xz
  y -> xy
  y -> yz
  z -> xz
  z -> yz
  xy -> xyz
  xz -> xyz
  yz -> xyz
}
%%%

%%

#Hasse diagrams (encore)

Is it possible to draw a Hasse diagram for any poset?

%%hidden(Show solution):
Note that our description of Hasse diagrams made use of the covers relation $\prec$. The covers relation, however, is not adequate to describe the structure of many posets. Consider the poset $\langle \mathbb R, \leq \rangle$ of the real numbers ordered by the standard comparison $\leq$. We have $0 < 1$, but how would we convey that with a Hasse diagram? The problem is that $0$ has no covers, even though it is not a maximal element in $\mathbb R$. In fact, for any $x \in \mathbb R$ such that $x > 0$, we can find a $y \in \mathbb R$ such that $0 < y < x$. This "infinite density" of $\mathbb R$ makes it impossible to depict using a Hasse diagram.
%%


Cauchy's theorem on subgroup existence

Cauchy's theorem states that if $G$ is a finite [-3gd] and $p$ is a [4mf prime] dividing $|G|$ the [3gg order] of $G$, then $G$ has a subgroup of order $p$. Such a subgroup is necessarily [47y cyclic] ([4jh proof]).

# Proof

The proof involves basically a single magic idea: from thin air, we pluck the definition of the following set.

Let $$X = \{ (x_1, x_2, \dots, x_p) : x_1 x_2 \dots x_p = e \}$$ the collection of $p$-[-tuple]s of elements of the group such that the group operation applied to the tuple yields the identity.
Observe that $X$ is not empty, because it contains the tuple $(e, e, \dots, e)$.

Now, the cyclic group $C_p$ of order $p$ [3t9 acts] on $X$ as follows: $$(h, (x_1, \dots, x_p)) \mapsto (x_2, x_3, \dots, x_p, x_1)$$ where $h$ is the generator of $C_p$.
So a general element $h^i$ acts on $X$ by sending $(x_1, \dots, x_p)$ to $(x_{i+1}, x_{i+2} , \dots, x_p, x_1, \dots, x_i)$.

This is indeed a group action (exercise).

%%hidden(Show solution):

- It certainly outputs elements of $X$, because if $x_1 x_2 \dots x_p = e$, then $$x_{i+1} x_{i+2} \dots x_p x_1 \dots x_i = (x_1 \dots x_i)^{-1} (x_1 \dots x_p) (x_1 \dots x_i) = (x_1 \dots x_i)^{-1} e (x_1 \dots x_i) = e$$
- The identity acts trivially on the set: since rotating a tuple round by $0$ places is the same as not permuting it at all.
- $(h^i h^j)(x_1, x_2, \dots, x_p) = h^i(h^j(x_1, x_2, \dots, x_p))$ because the left-hand side has performed $h^{i+j}$ which rotates by $i+j$ places, while the right-hand side has rotated by first $j$ and then $i$ places and hence $i+j$ in total.
%%

Now, fix $\bar{x} = (x_1, \dots, x_p) \in X$.

By the [4l8 Orbit-Stabiliser theorem], the [4v8 orbit] $\mathrm{Orb}_{C_p}(\bar{x})$ of $\bar{x}$ divides $|C_p| = p$, so (since $p$ is prime) it is either $1$ or $p$ for every $\bar{x} \in X$.

Now, what is the size of the set $X$?
%%hidden(Show solution):
It is $|G|^{p-1}$.

Indeed, a single $p$-tuple in $X$ is specified precisely by its first $p$ elements; then the final element is constrained to be $x_p = (x_1 \dots x_{p-1})^{-1}$.
%%

Also, the orbits of $C_p$ acting on $X$ partition $X$ ([4mg proof]).
Since $p$ divides $|G|$, we must have $p$ dividing $|G|^{p-1} = |X|$.
Therefore since $|\mathrm{Orb}_{C_p}((e, e, \dots, e))| = 1$, there must be at least $p-1$ other orbits of size $1$, because each orbit has size $p$ or $1$: if we had fewer than $p-1$ other orbits of size $1$, then there would be at least $1$ but strictly fewer than $p$ orbits of size $1$, and all the remaining orbits would have to be of size $p$, contradicting that $p \mid |X|$.
[todo: picture of class equation]

Hence there is indeed another orbit of size $1$; say it is the singleton $\{ \bar{x} \}$ where $\bar{x} = (x_1, \dots, x_p)$.

Now $C_p$ acts by cycling $\bar{x}$ round, and we know that doing so does not change $\bar{x}$, so it must be the case that all the $x_i$ are equal; hence $(x, x, \dots, x) \in X$ and so $x^p = e$ by definition of $X$.


Equaliser (category theory)

In [-4c7], an *equaliser* of a pair of arrows $f, g: A \to B$ is an object $E$ and a universal arrow $e: E \to A$ such that $ge = fe$.
Explicitly, $ge = fe$, and for any object $X$ and arrow $x: X \to A$ such that $fx = gx$, there is a unique factorisation $\bar{x} : X \to A$ such that $e \bar{x} = x$.


Orbit-stabiliser theorem

[summary:
A [-3t9 group action] of a group $G$ acting on a set $X$ describes how $G$ sends elements of $X$ to other elements of $X$.  Given a specific element $x \in X$, the [-4mz stabiliser] is all those elements of the group which send $x$ back to itself, and the [-4v8 orbit] of $x$ is all the elements to which $x$ can get sent.

This theorem tells you that $G$ is divided into equal-sized pieces using $x$. Each piece "looks like" the stabilizer of $x$ (and is the same size), and the orbit of $x$ tells you how to "move the piece around" over $G$ in order to cover it.

Put another way, each element $y$ in the orbit of $x$ is transformed "in the same way" by $G$ relative to $y$.

This theorem is closely related to [-4jn Lagrange's Theorem].
]

[summary(Technical): Let $G$ be a finite [-3gd], [3t9 acting] on a set $X$. Let $x \in X$.
Writing $\mathrm{Stab}_G(x)$ for the [4mz stabiliser] of $x$, and $\mathrm{Orb}_G(x)$ for the [4v8 orbit] of $x$, we have $$|G| = |\mathrm{Stab}_G(x)| \times |\mathrm{Orb}_G(x)|$$ where $| \cdot |$ refers to the size of a set.]

Let $G$ be a finite [-3gd], [3t9 acting] on a set $X$. Let $x \in X$.
Writing $\mathrm{Stab}_G(x)$ for the [4mz stabiliser] of $x$, and $\mathrm{Orb}_G(x)$ for the [4v8 orbit] of $x$, we have $$|G| = |\mathrm{Stab}_G(x)| \times |\mathrm{Orb}_G(x)|$$ where $| \cdot |$ refers to the size of a set.

This statement generalises to infinite groups, where the same proof goes through to show that there is a [499 bijection] between the [4j4 left cosets] of the group $\mathrm{Stab}_G(x)$ and the orbit $\mathrm{Orb}_G(x)$.

# Proof

Recall that the [-4lt] of the parent group.

Firstly, it is enough to show that there is a bijection between the left cosets of the stabiliser, and the orbit.
Indeed, then $$|\mathrm{Orb}_G(x)| |\mathrm{Stab}_G(x)| = |\{ \text{left cosets of} \ \mathrm{Stab}_G(x) \}| |\mathrm{Stab}_G(x)|$$
but the right-hand side is simply $|G|$ because an element of $G$ is specified exactly by specifying an element of the stabiliser and a coset.
(This follows because the [4j5 cosets partition the group].)

## Finding the bijection

Define $\theta: \mathrm{Orb}_G(x) \to \{ \text{left cosets of} \ \mathrm{Stab}_G(x) \}$, by $$g(x) \mapsto g \mathrm{Stab}_G(x)$$

This map is well-defined: note that any element of $\mathrm{Orb}_G(x)$ is given by $g(x)$ for some $g \in G$, so we need to show that if $g(x) = h(x)$, then $g \mathrm{Stab}_G(x) = h \mathrm{Stab}_G(x)$.
This follows: $h^{-1}g(x) = x$ so $h^{-1}g \in \mathrm{Stab}_G(x)$.

The map is [4b7 injective]: if $g \mathrm{Stab}_G(x) = h \mathrm{Stab}_G(x)$ then we need $g(x)=h(x)$.
But this is true: $h^{-1} g \in \mathrm{Stab}_G(x)$ and so $h^{-1}g(x) = x$, from which $g(x) = h(x)$.

The map is [4bg surjective]: let $g \mathrm{Stab}_G(x)$ be a left coset.
Then $g(x) \in \mathrm{Orb}_G(x)$ by definition of the orbit, so $g(x)$ gets taken to $g \mathrm{Stab}_G(x)$ as required.

Hence $\theta$ is a well-defined bijection.


Needs exercises

One way the user can make sure they've understood the material is to do [4n2 exercises]. They should usually be on their own [17b lens], as additional "homework problems" that users can solve to get more practice. They can also be embedded in the explanation pages, so the user can solve them as they are learning, especially on [1rt path] pages where the reader has asked for exercises.

Many pages could be improved by adding exercises, but priority should be given to:

* Pages which don't currently have exercises.
* Pages which have a full explanation of a concept (rather than just [4gs defienitions]).
* Pages which are a requisite that many other concepts rely on.
* Pages which are mostly self-contained, rather than overviews of a large and multifaceted concept (exercises should go with child pages explaining the subtopics).


Join and meet: Exercises

Try these exercises to test your knowledge of joins and meets.

Tangled up
--------------------

![a big crazy poset](http://i.imgur.com/cYqkKm7.png)

Determine whether or not the following joins and meets exist in the poset depicted by the above Hasse diagram.

$c \vee b$

%%hidden(Show solution):
This join does *not* exist, because $i$ and $j$ are incomparable upper bounds of $\{ c, b \}$, and no smaller upper bounds of $\{ c, b \}$ exist.
%%

$g \vee e$

%%hidden(Show solution):
This join exists. $g \vee e = j$.
%%

$j \wedge f$

%%hidden(Show solution):
This meet does *not* exist, because $d$ and $b$ are incomparable lower bounds of $\{ j, f \}$, and no larger lower bounds of $\{ j, f \}$ exist.
%%

$\bigvee \{a,b,c,d,e,f,g,h,i,j,k,l\}$

%%hidden(Show solution):
This join exists. It is $l$.
%%

Join fu
------------

Let $P$ be a poset, $S \subseteq P$, and $p \in P$. Prove that if both $\bigvee S$ and $(\bigvee S) \vee p$ exist then $\bigvee (S \cup \{p\})$ exists as well, and $(\bigvee S) \vee p = \bigvee (S \cup \{p\})$.

%%hidden(Show solution):
For any $X \subset P$, let $X^U$ denote the set of upper bounds of $X$. The above proposition follows from the fact that $\{\bigvee S, p\}^U = (S \cup p)^U$, which is apparent from the following chain of bi-implications:

$q \in \{\bigvee S, p\}^U \iff$

for all $s \in S, q \geq \bigvee S \geq s$, and $q \geq p \iff$

$q \in (S \cup \{p\})^U$.

If two subsets of a poset have the same set of upper bounds, then either they both lack a least upper bound, or both have the same least upper bound.
%%

Meet fu
------------

Let $P$ be a poset, $S \subseteq P$, and $p \in P$. Prove that if both $\bigwedge S$ and $(\bigwedge S) \wedge p$ exist then $\bigwedge (S \cup \{p\})$ exists as well, and $(\bigwedge S) \wedge p = \bigwedge(S \cup \{p\})$.

%%hidden(Show solution):
Note that the proposition we are trying to prove here is the dual of the one stated in join fu. Thanks to the duality principle, this theorem therefore comes for free with our solution to join fu.
%%

Quite a big join
--------------------------

In the poset $\langle \mathbb N, | \rangle$ discussed in [43s], does $\bigvee \mathbb N$ exist? If so, what is it?


Stabiliser is a subgroup

[summary: Given a [-3gd group] $G$ [-3t9 acting] on a [-3jz set] $X$, the [-4mz stabiliser] of some [-element_mathematics element] $x \in X$ is a [-subgroup subgroup] of $G$.
]

Let $G$ be a [-3gd] which [3t9 acts] on the [-3jz set] $X$.
Then for every $x \in X$, the [4mz stabiliser] $\mathrm{Stab}_G(x)$ is a [-subgroup subgroup] of $G$.

# Proof

We must check the group axioms.

- The [-54p identity], $e$, is in the stabiliser because $e(x) = x$; this is part of the definition of a group action.
- [-3gy Closure] is satisfied: if $g(x) = x$ and $h(x) = x$, then $(gh)(x) = g(h(x))$ by definition of a group action, but that is $g(x) = x$.
- [-3h4 Associativity] is inherited from the parent group.
- [-inverse_mathematics Inverses]: if $g(x) = x$ then $g^{-1}(x) = g^{-1} g(x) = e(x) = x$.


Prime number

A [-45h] $n > 1$ is *prime* if it has no [divisor_number_theory divisors] other than itself and $1$.
Equivalently, it has the property that if $n \mid ab$ %%note:That is, $n$ divides the product $ab$%% then $n \mid a$ or $n \mid b$.
Conventionally, $1$ is considered to be neither prime nor [composite_number composite] (i.e. non-prime).

# Examples

- The number $2$ is prime, because its divisors are $1$ and $2$; therefore it has no divisors other than itself and $1$.
- The number $3$ is also prime, as are $5, 7, 11, 13, \dots$.
- The number $4$ is not prime; neither are $6, 8, 9, 10, 12, \dots$.

# Properties

- There are infinitely many primes. ([54r Proof.])
- Every natural number may be written as a product of primes; moreover, this can only be done in one way (if we count "the same product but with the order swapped" as being the same: for example, $2 \times 3 = 3 \times 2$ is just one way of writing $6$). ([fundamental_theorem_of_arithmetic Proof.])

# How to find primes

If we want to create a list of all the primes below a given number, or the first $n$ primes for some fixed $n$, then an efficient way to do it is the [sieve_of_eratosthenes Sieve of Eratosthenes].
(There are other sieves available, but Eratosthenes is the simplest.)

There are many [primality_testing tests] for primality and for compositeness.

# More general concept

This definition of "prime" is, in a more general [3gq ring-theoretic] setting, known instead as the property of [5m1 irreducibility].
Confusingly, there is a slightly different notion in this ring-theoretic setting, which goes by the name of "prime"; this notion has [prime_element_ring_theory a separate page on Arbital].
In the ring of integers, the two ideas of "prime" and "irreducible" actually coincide, but that is because the integers form a ring with several very convenient properties: in particular, being a [euclidean_domain Euclidean domain], they are a [-principal_ideal_domain] (PID), and [pid_implies_ufd PIDs have unique factorisation].

[todo: add requisite for divisor_number_theory]


Group orbits partition

Let $G$ be a [-3gd], [3t9 acting] on the set $X$.
Then the [group_orbit orbits] of $X$ under $G$ form a [set_partition partition] of $X$.

# Proof

We need to show that every element of $X$ is in an orbit, and that if $x \in X$ lies in two orbits then they are the same orbit.

Certainly $x \in X$ lies in an orbit: it lies in the orbit $\mathrm{Orb}_G(x)$, since $e(x) = x$ where $e$ is the identity of $G$.
(This follows by the definition of an action.)

Suppose $x$ lies in both $\mathrm{Orb}_G(a)$ and $\mathrm{Orb}_G(b)$, where $a, b \in X$.
Then $g(a) = h(b) = x$ for some $g, h \in G$.
This tells us that $h^{-1}g(a) = b$, so in fact $\mathrm{Orb}_G(a) = \mathrm{Orb}_G(b)$; it is an exercise to prove this formally.

%%hidden(Show solution):
Indeed, if $r \in \mathrm{Orb}_G(b)$, then $r = k(b)$, say, some $k \in G$.
Then $r = k(h^{-1}g(a)) = kh^{-1}g(a)$, so $r \in \mathrm{Orb}_G(a)$.

Conversely, if $r \in \mathrm{Orb}_G(a)$, then $r = m(b)$, say, some $m \in G$.
Then $r = m(g^{-1}h(b)) = m g^{-1} h (b)$, so $r \in \mathrm{Orb}_G(b)$.
%%


Product (Category Theory)

This simultaneously captures the concept of a product of [-3jz sets], [-3rb posets], [-3gd groups], [-topological_space topological spaces] etc. In addition, like any [-universal_property universal construction], this characterization does not differentiate between [-4f4 isomorphic] versions of the product, thus allowing one to abstract away from an arbitrary, [-specific_construction_category_theory specific construction].

##Definition##
Given a pair of objects $X$ and $Y$ in a category $\mathbb{C}$, the **product** of $X$ and $Y$ is an object $P$ *along with a pair of morphisms* $f: P \rightarrow X$ and $g: P \rightarrow Y$ satisfying the following [-universal_property universal] condition:

Given any other object $W$ and morphisms $u: W \rightarrow X$ and $v:W \rightarrow Y$ there is a *unique* morphism $h: W \rightarrow P$ such that $fh = u$ and $gh = v$.


Empirical probabilities are not exactly 0 or 1

[summary:  [Cromwell's Rule](https://en.wikipedia.org/wiki/Cromwell%27s_rule) in statistics forbids us to assign probabilities of exactly $0$ or $1$ to any empirical proposition - that is, it is always possible to be mistaken.

- Probabilities of exactly $0$ or $1$ correspond to infinite [1zh log odds], meaning that no finite amount of evidence can ever suffice to reach them, or overturn them.  Once you assign probability $0$ or $1,$ you can never change your mind.
- Sensible [4mr universal priors] never assign probability exactly $0$ or $1$ to any predicted future observation - their hypothesis space is always broad enough to include a scenario where the future is different from the past.]

[Cromwell's Rule](https://en.wikipedia.org/wiki/Cromwell%27s_rule) in statistics argues that no empirical proposition should be assigned a subjective probability of *exactly* $0$ or $1$ - it is always *possible* to be mistaken.  (Some argue that this rule should be generalized to logical facts as well.)

A probability of exactly $0$ or $1$ corresponds to infinite [1zh log odds], and would require infinitely [22x strong] evidence to reach starting from any finite [1rm prior].  To put it another way, if you don't start out infinitely certain of a fact before making any observations (before you were born), you won't reach infinite certainty after any finite number of observations involving finite probabilities.

All sensible [4mr universal priors] seem so far to have the property that they never assign probability exactly $0$ or $1$ to any predicted future observation, since their hypothesis space is always broad enough to include an imaginable state of affairs in which the future is different from the past.

If you did assign a probability of exactly $0$ or $1,$ you would be unable to [1ly update] no matter how much contrary evidence you observed.  [1rb Prior odds] of 0 : 1 (or 1 : 0), times any finite [1rq likelihood ratio], end up yielding 0 : 1 (or 1 : 0).

As Rafal Smigrodski put it:

> "I am not totally sure I have to be always unsure.  Maybe I could be legitimately sure about something.  But once I assign a probability of 1 to a proposition, I can never undo it.  No matter what I see or learn, I have to reject everything that disagrees with the axiom.  I don't like the idea of not being able to change my mind, ever."


Universal prior

A "universal" [27p prior] is a [3tb probability distribution] that assigns positive probability to *every thinkable hypothesis*, for some reasonable meaning of "every thinkable hypothesis".  A central example would be [11w Solomonoff induction], in which the observations are a sequence of bits, the universal prior is "every possible computer program that generates bit sequences", and every such computer program starts with a positive probability.


Stabiliser (of a group action)

[summary: The stabilizer of an element $x$ under the action of a group $G$ is the set ([4lt actually a subgroup]) of elements of $G$ which leave $x$ unchanged. ]

Let the [-3gd] $G$ [3t9 act] on the set $X$.
Then for each element $x \in X$, the *stabiliser* of $x$ under $G$ is $\mathrm{Stab}_G(x) = \{ g \in G: g(x) = x \}$.
That is, it is the collection of elements of $G$ which do not move $x$ under the action.

The stabiliser of $x$ is a subgroup of $G$, for any $x \in X$. ([4lt Proof.])

A closely related notion is that of the [4v8 orbit] of $x$, and the very important [4l8 Orbit-Stabiliser theorem] linking the two.


Exercise

One way readers can reinforce or make sure they've understood a concept is to do exercises. These pages give a collection of questions for this purpose.


Why is the decimal expansion of log2(3) infinite?

[summary:
It takes more than one but less than two [binary_digit binary digits] to encode a [4sj 3-digit], so $\log_2(3)$ must be between 1 and 2. ([427 Wait, what?]). It takes more than 15 but less than 16 binary digits to encode ten 3-digits, so $10 \cdot \log_2(3)$ must be between 15 and 16, which  means $1.5 < \log_2(3) < 1.6.$ It takes more than 158 but less than 159 binary digits to encode a hundred 3-digits, so $1.58 < \log_2(3) < 1.59.$ And so on. Because no power of 3 is ever equal to any power of 2, $10^n \cdot \log_2(3)$ will never quite be a whole number, no matter how large $n$ is.]

$\log_2(3)$ starts with

1.5849625007211561814537389439478165087598144076924810604557526545410982277943585625222804749180882420909806624750591673437175524410609248221420839506216982994936575922385852344415825363027476853069780516875995544737266834624612364248850047581810676961316404807130823233281262445248670633898014837234235783662478390118977006466312634223363341821270106098049177472541357330110499026268818251703576994712157113638912494135752192998699040767081539505404488360

and goes on indefinitely. Why is it 1.58... in particular? Well, it takes more than one but less than two [binary_digit binary digits] to encode a [4sj 3-digit], so $\log_2(3)$ must be between 1 and 2. ([427 Wait, what?]). It takes more than 15 but less than 16 binary digits to encode ten 3-digits, so $10 \cdot \log_2(3)$ must be between 15 and 16, which  means $1.5 < \log_2(3) < 1.6.$ It takes more than 158 but less than 159 binary digits to encode a hundred 3-digits, so $1.58 < \log_2(3) < 1.59.$ And so on. Because no power of 3 is ever equal to any power of 2, $10^n \cdot \log_2(3)$ will never quite be a whole number, no matter how large $n$ is.

Thus, $\log_2(3)$ has no finite decimal expansion, because $3$ is not a [4zq rational] [-power] of $2$. Using this argument, we can see that $\log_b(x)$ is an integer if (and only if) $x$ is a power of $b$, and that $\log_b(x)$ only has a finite expansion if some power of $x$ is a power of $b.$


Guide

Meta tag for the start page of a [327 multi-page guide].


Needs brief summary

Pages which have a long or in-depth main summary should also have a brief summary. These show up as an alternate tab on the [1kl summary popover] when a user hovers over a [17f greenlink]. Add this tag to pages that need to have a brief summary.

Pages may also require a [433 general] or [4pt technical] summary.


n-digit

An $n$-digit is a physical object that can be stably placed into any of $n$ distinguishable states. For example, a coin (which can be placed heads or tails) and a single bit of memory on a computer (which either has a high volt level or a low volt level) are both examples of 2-digits. A [-42d] is an example of a 10-digit. One die is an example of a 6-digit; two dice together are an example of a 36-digit (because they can be placed in 36 different ways).

What does and doesn't count as an $n$-digit depends on context and convention: For example, if you want to communicate a message to me by placing a penny heads-side up and choosing whether to point Abraham Lincoln's face either north, south, east, or west, then, for the purposes of the two of us, that penny is a 4-digit rather than a 2-digit. The definition of "stably placed" is also a bit up-for-grabs: If you're writing a computer program and need to store a [3v9 256-message] in short-term memory, then a byte of RAM will do, but if you need to store the same 256-message for a long period of time, you may need to use a less temporary 256-digit (such as a hard drive).

Note that it's possible to emulate $m$-digits using $n$-digits, in general. If $m < n$ then an $n$-digit is trivially an $m$-digit (i.e., you can use a digit wheel like a 7-digit in a pinch), and if $m > n$ then, given enough $n$-digits, you can make do. For example, 3 coins can be used to encode an 8-digit. See also [emulating_digits].


Emulating digits

In general, given enough $n$-digits, you can emulate an $m$-digit, for any $m, n \in$ [45h $\mathbb N$]. If $m < n,$ you can emulate an $m$-digit using just one $n$-digit — in other words, you can use a [-42d] like a $7$-digit if you want to, by just ignoring three of the possible ways to set the digit wheel. If $m > n,$ things are a bit more difficult, but only slightly.

Basically, with 2 $n$-digits, you can emulate a $n^2$-digit, as follows. Using your two $n$-digits, encode a number $(x, y)$ where $0 \le x < n$ and $0 \le y < n$. Interpret $(x, y)$ as $xn + y.$ You have now encoded a number between 0 (if  $x = y = 0$) and $n^2 - 1$ (if $x = y = n-1$). Congratulations, you just used two $n$-digits to make an $n^2$ digit!

You can use the same strategy to emulate $n^3$-digits (interpret $(x, y, z)$ as $xn^2 + yn + z$), $n^4$-digits (you get the picture), and so on. Now, to emulate an $m$-digit, just pick an exponent $a$ such that $n^a > m,$ collect $a$ copies of an $n$-digit, and you're done.

This isn't necessarily the most efficient way to use $n$-digits to encode $m$-digits. For example, if $m$ is 1,000,001 and $n$ is 10, then you need seven 10-digits. Seven 10-digits are enough to emulate a 10-million-digit, whereas $m$ is a mere million-and-one-digit — paying for a 10-million-digit when all you needed was an $m$-digit seems a bit excessive. For some different methods you can use to recover your losses when encoding one type of digit using another type of digit, see [44l] and [3ty]. (These techniques are fairly useful in practice, given that modern computers encode everything using [3p0 bits], i.e. 2-digits, and so it's useful to know how to efficiently encode $m$-messages using bits when $m$ is pretty far from the nearest power of 2.)


Decimal notation

Seventeen is the number that represents as many things as there are x marks at the end of this sentence: xxxxxxxxxxxxxxxxx. Writing out numbers by saying "the number representing how many things there are in this pile:" gets unwieldy when the pile gets large. Thus, we _represent_ numbers using the [numeral numerals] 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Specifically, we write the number representing this many things: xxx as "3", and the number representing this many things: xxxxxxxxxxx as 11, and the number seventeen as "17". This is called "decimal notation,"  because there are ten different symbols that we use. Numbers don't have to be written down in decimal notation, it's also possible to write them down in other notations such as [-binary_notation]. Some numbers can't even be written out in decimal notation (in full); consider, for example, the number $e$ which, in decimal notation, starts out with the digits 2.71828... and just keeps going.

# How decimal notation works

How do you know that 17 is the number that represents the number of xs in this sequence: xxxxxxxxxxxxxxxxx? In practice, you know this because the rules of decimal notation were ingrained in you in a young child. But do you know those rules explicitly? Could you write out a series of rules for taking in some input symbols like '2', '4', and '6' and using those to figure out how many pebbles to add to a pile?

The answer, of course, is this many:

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

But how do we perform that conversion in general?

In short, the number 246 represents $(2 \cdot 100) + (4 \cdot 10) + (6 \cdot 1),$ so as long as we know how to do addition and multiplication, and as long as we know what the basic numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 mean, and as long as we know how to get to [power powers] of 10 (1, 10, 100, 1000, ...), then we can explicitly understand decimal notation.

(What do the basic numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 mean? By convention, they represent as many things as are in the following ten sequences of xs: , x, xx, xxx, xxxx, xxxxx, xxxxxx, xxxxxxx, xxxxxxxx, and xxxxxxxxx, respectively.)

This explanation assumes that you're already quite familiar with decimal notation. Explaining decimal notation from scratch to someone who doesn't already know it (which was a task people _actually had to do_ back when half the world was using Roman numerals, a much less convenient system for representing numbers) is a fun task; to see what that looks like, refer to [+representing_numbers_from_scratch].

# Other common notations

The above text made use of [-unary_notation], which is a method of representing numbers by making a number of marks that correspond to the represented number. For example, in unary notation, 17 is written xxxxxxxxxxxxxxxxx (or ||||||||||||||||| or whatever, the actual marks don't matter). This is perhaps somewhat easier to understand, but writing large numbers like 93846793284756 gets rather ungainly.

Historical notations include [https://en.wikipedia.org/wiki/Roman_numerals Roman numerals], which were a pretty bad way to represent numbers. (It took humanity quite some time to find _good_ tools for representing numbers; the decimal notation that's been ingrained in your head since early childhood is the result of many centuries worth of effort. It's much harder to invent good representations of numbers when you don't even have good tools for writing down and reasoning about numbers. Furthermore, the modern tools for representing numbers aren't necessarily ideal!)

Common notations in modern times (aside from decimal notation) include [-binary_notation] (often used by computers), [-hexadecimal_notation] (which is a useful format for humans reading binary notation). Binary notation and hexadecimal notation are very similar to decimal notation, with the difference that binary uses only two distinct symbols (instead of ten), and hexadecimal uses sixteen.


Inverse function

If a function $g$ is the inverse of a function $f$, then $g$ undoes $f$, and $f$ undoes $g$. In other words, $g(f(x)) = x$ and $f(g(y)) = y$. An inverse function takes as its [3js domain] the [3lm range] of the original function, and the [3lm range] of the inverse function is the [3js domain] of the original. To put that another way, if $f$ maps $A$ onto $B$, then $g$ maps $B$ back onto $A$. To indicate the inverse of a function $f$, we write $f^{-1}$.

## Examples ##

$$y=f(x) = x^3\ \ \ \ \ \ \ \ \ \ \ \   x=f^{-1}(y) = y^{1/3}$$
$$y=f(x) = e^x\ \ \ \ \ \ \ \ \ \ \ \   x=f^{-1}(y) = ln(y)$$
$$y=f(x) = x+4\ \ \ \ \ \ \ \ \ \ \ \  x=f^{-1}(y) = y-4$$


Quotient group

[summary:
Given a [-3gd group] $G$ with operation $\bullet$ and a special kind of [-subgroup subgroup] $N \leq G$ called the "[-4h6 normal subgroup]", there is a way of "dividing" $G$ by $N$, written $G/N$. Usually this is defined so that each [-element_mathematics element] of $G/N$ is a [-subset subset] of $G$. In fact, each of them is an [-equivalence_class equivalence class] of elements in $G$, so called because we think of everything in one equivalence class as being equivalent. One of the equivalence classes is $N$, and each of the other equivalence classes can be seen as $N$ "shifted around" inside $G$. Because they are equivalence classes, each element of $G$ only occurs in exactly one of them.

This collection $G/N$ of equivalence classes has a [-group_structure group structure] based on the structure on $G$. Write the [-group_operation operation] as $\circ$. Say $A$ and $B$ are two equivalence classes in $G/N$. Then $A \circ B$ is defined by taking an element $a \in A$ and $b \in B$, multiplying them as $a \bullet b$ and checking in which equivalence class this product ends up.

There are some things to check, namely that the equivalence class in which we end up is the same regardless of which elements in $A$ and $B$ we happen to pick. This is guaranteed by the fact that $N$ is normal in $G$.

There is also a [-47t group homomorphism] $\phi$ from $G$ to $G/N$ that "collapses" the elements by sending each element in $G$ to the equivalence class in which it lies. This is called collapsing because lots of elements in $G$ are sent to the same element in $G/N$, so we can see $G/N$ as a collapsed version of $G$.
]

[summary(brief):
A **quotient group** $G/N$ of a [-3gd group] $G$ by a [-4h6 normal subgroup] $N$ is obtained by dividing up the group into pieces ([-equivalence_class equivalence classes]), and then treating everything in one class the same way (by treating each class as a single element). The quotient group has a group structure defined on it based on the original structure of $G$, that works 'basically the same as $G$ up to equivalence'.
 ]

[summary(technical):
Given a [-3gd group] $(G, \bullet)$ and a [-4h6 normal subgroup] $N \unlhd G$. The **quotient** of $G$ by $N$, written $G/N$, has as [-3gz underlying set] the set of (left)-[-4j4 cosets] of $N$ in $G$ and as operation $\circ$ which is defined as $aN \circ  bN = (a \bullet b) N$, where $xN = \{xn : n \in N\}$ for each $x \in G$.

The operation $\circ$ is [well_defined well defined] in the sense that if other representatives $a'$ and $b'$ are chosen such that
$a'N = aN$ and $b'N = bN$ then also $(a' \bullet b')N = (a \bullet b)N$.

There is a [-canonical canonical] [-47t homomorphism] (sometimes called the [-quotient_projection projection]) $\phi: G \rightarrow G/N: a \mapsto aN$.

This is a special case of a [-quotient_universal_algebra quotient] from universal algebra.
]

[summary(examples):
Given the [-3gd group] of [-48l integers] $\mathbb{Z}$, and the [-nomral_subgroup normal subgroup] $2 \mathbb{Z}$ of all even numbers, we can form the group $\mathbb{Z}/2\mathbb{Z}$. This group has only two [-element elements] and [-personification_in_mathematics only cares] if a number is odd or even. It tells us that the sum of an odd and an even number is odd, that the sum of two even numbers is even, and the sum of two odd numbers is also even!
]

[summary(motivation):
Let's say we have a group. Maybe the group is kinda large and unwieldy, and we want to find an easier way to think about it. Or maybe we just want to focus on a certain aspect of the group. Some of the actions will change things in ways we just don't really care about, or don't mind ignoring for now. So let's create a group homomorphism that will map all these actions to the identity action in a new group. The image of this homomorphism will be a group much like the first, except that it will ignore all the effects that come from those actions that we're ignoring - just what we wanted! This new group is called the quotient group.
]

#The basic idea
Let's say we have a [3gd group]. Maybe the group is kinda large and unwieldy, and we want to find an easier way to think about it. Or maybe we just want to focus on a certain aspect of the group. Some of the actions will change things in ways we just don't really care about, or don't mind ignoring for now. So let's create a [47t group homomorphism] that will map all these actions to the identity action in a new group. The image of this homomorphism will be a group much like the first, except that it will ignore all the effects that come from those actions that we're ignoring - just what we wanted! This new group is called the quotient group.

#Definition

We start with our group $G$. The actions we want to ignore form a group $N$, which must be a [4h6 normal subgroup] of $G$. The quotient group is then called $G/N$, and has a canonical homomorphism $\phi: G \rightarrow G/N$ which maps $g \in G$ to the [4j4 coset] $gN$.

## The divisor group
In the definition, we require the divisor $N$, to be a normal subgroup of $G$. Why? Well first, let's see why requiring $N$ to be a group makes sense. Remember that $N$ has the actions whose effects we want to ignore. So it makes sense that it should contain the identity action, which has no effect. It also is reasonable that it would be closed under the group operation - doing two things we don't care about shouldn't change anything we care about. Together, these two properties imply it is a [subgroup subgroup]: $N \le G$.

A subgroup is great, but it isn't quite good enough by itself to work here. That's because we want the quotient group to preserve the overall structure of the group, i.e. it should preserve the group multiplication. In other words, there needs to be a [47t group homomorphism] $\phi$ from $G$ to $G/N$. Since $N$ is the subgroup of things we want to ignore, all its actions should get mapped to the identity action under this homomorphism. That means it's the [49y kernel] of the homomorphism $\phi$, which means it's a [4h6 normal subgroup]: $N \trianglelefteq G$.

## Cosets
What exactly are the elements of the new group? They are [equivalence_class equivalence classes] of actions, the sets $gN = \{gn : n \in N\}$ where $g \in G$, also known as a [4j4 coset]. The identity element is the set $N$ itself. Multiplication is defined by $g_1N \cdot g_2N = (g_1g_2)N$.

# Generalizes the idea of a quotient
What gives a quotient group the right to call itself a quotient? If $G$ and $N$ both have finite order, then $|G/N| = |G|/|N|$, which can be proved by the fact that $G/N$ consists of the cosets of $N$ in $G$, and that these [4j8 cosets are the same size], and [4j5 partition] $G$.

#Example
Suppose you have a collection of objects, and you need to split them into two equal groups. So you are trying to determine under what circumstances changing the number of objects will affect this property. You notice that changing the size of the collection by certain numbers such as 0, 2, 4, 24, and -6 doesn't affect this property.

The set of different size changes can be modeled as the additive group of integers $\mathbb Z$. The changes that don't affect this property also form a group: $2\mathbb Z = \{2n : n\in \mathbb Z\}$. Exercise: verify that this is a normal subgroup of $\mathbb Z$.

This subgroup gives us two cosets: $0 + 2\mathbb Z$ and $1 + 2\mathbb Z$ (remember that $+$ is the group operation in this example), which are the elements of our quotient group. We will give them their conventional names: $\text{even}$ and $\text{odd}$, and we can apply the coset multiplication rule to see that $\text{even}+ \text{even} = \text{even}$, $\text{even} + \text{odd} = \text{odd}$, and $\text{odd} + \text{odd} = \text{even}$.

Instead of thinking about specific numbers, and how they will change our ability to split our collection of objects into two equal groups, we now have reduced the problem to its essence. Only the parity matters, and it follows the simple rules of the quotient group we discovered.

#See also

 - [4jn Lagrange's theorem].
 - [first_isomorphism_theorem The first isomorphism theorem].


Exponential

[summary: The exponent base $b$ of a number $x$, written $b^x,$ is what you get when you multiply 1 by $b$, $x$ times. For example, the exponent base 10 of 3 is $10^3$ and pronounced "10 to the power 3" is 1000, because $10 \cdot 10 \cdot 10 = 1000$. Similarly, $2^4=16,$ because $2 \cdot 2 \cdot 2 \cdot 2 = 16.$

The input $x$ may be fractional. For example, $10^{1/2}$ is the number you get when you multiply 1 by 10 half a time, where "multiplying by 10 half a time" means multiplying by a number $n$ such that if you multiplied by $n$ twice, you would have multiplied by 10 once. In this case, $n \approx 3.16,$ because then $n \cdot n \approx 10.$]

[todo: Add numerous other summaries. See, e.g., the summaries at /p/logarithm.]
[todo: I suggest making the main page for exponential mirror the main page at /p/logarithm.]

An exponential is a [-3jy] that can be represented by some [-constant] taken to the [power_mathematics power] of a [-variable]. The name comes from the fact that the variable is the *exponent* of the expression.

## Exponential Growth

Exponentials are most useful in describing growth patterns where the growth rate is [4w3 proportional] to the amount of the thing that's growing. They can be represented by the formula: $f(x) = c \times a^x$, where $c$ is the starting value and $a$ is the growth factor.

The classic example of exponential growth is [-compound_interest]. If you have \$100 in a bank account that gives you 2% interest every year, then every year your money is multiplied by $1.02$. This means you can represent your account balance as $f(x) = 100 \times 1.02^x$, where $x$ is the number of years your money has been in the account.

Another example is a dividing cell. If one cell is placed into an infinite culture and splits once every hour, the number of cells in the culture after $x$ hours is $f(x) = 1 \times 2^x$ (assuming none of the cells die).

### Recursive definition

We mentioned earlier that in an exponential function, the growth rate is proportional to the amount of the thing that's growing. In the compound interest example, we can write each value in terms of the previous value: $f(x) = f(x-1) \times 1.02$. Therefore, the amount of growth at every step can be represented as $\Delta f(x) = f(x+1) - f(x) = 0.02 \times f(x)$.

This makes exponential growth a *memoryless growth function*, as the growth rate depends only on current information. Compare this to [-simple_interest], where the interest only grows at a percentage of the initial value. Then we would have to write: $f(x) = f(x-1) + 0.02 \times f(0)$, and because $f(0)$ is constant while $f(x)$ continues to grow, we cannot express the growth rate in terms of only current information — we have to keep "in memory" the initial balance to calculate the growth rate.


Work in progress

A meta tag for pages unfinished pages which an author is still making major changes to.

[todo: Define/explain guidelines for use.]


Most complex things are not very compressible

Although the [46h halting problem] means we can't *prove* it doesn't happen, it would nonetheless be *extremely surprising* if some 100-state Turing machine turned out to print the exact text of Shakespeare's *Romeo and Juliet.*  Unless something was specifically generated by a simple algorithm, the Vast supermajority of data structures that *look* like they have high [5v algorithmic complexity] actually *do* have high algorithmic complexity.  Since there are at most $2^{101}$ programs that can be specified with at most 100 bits (in any particular language), we can't fit all the 1000-bit data structures into all the 100-bit programs.  While *Romeo and Juliet* is certainly highly compressible, relative to most random bitstrings of the same length, it would be shocking for it to compress *all the way down* to a 100-state Turing machine.  There just aren't enough 100-state Turing machines for one of their outputs to be *Romeo and Juliet*.  Similarly, if you start with a 10 kilobyte text file, and 7zip compresses it down to 2 kilobytes, no amount of time spent trying to further compress the file using other compression programs will ever get that file down to 1 byte.  For any given compressor there's at most 256 starting files that can ever be compressed down to 1 byte, and your 10-kilobyte text file almost certainly isn't one of them.

This takes on defensive importance with respect to refuting the probability-theoretic fallacy, "Oh, sure, Occam's Razor seems to say that this proposition is complicated.  But how can you be sure that this apparently complex proposition wouldn't turn out to be generated by some very simple mechanism?"  If we consider a [1rd partition] of 10,000 possible propositions, collectively having a 0.01% probability on average, then all the arguments in the world for why various propositions might have unexpectedly high probability, must still add up to an average probability of 0.01%.  It can't be the case that after considering that proposition 1 might have secretly high probability, and considering that proposition 2 might have secretly high probability, and so on, we end up assigning 5% probability to each proposition, because that would be a total probability of 500.  If we assign prior probabilities using an algorithmic-complexity Occam prior as in [11w Solomonoff induction], then the observation that "most apparently complex things can't be further compressed into an amazingly simple Turing machine", is the same observation as that "most apparently Occam-penalized propositions can't turn out to be simpler than they look" or "most apparently subjectively improbable things can't turn out to have unseen clever arguments that would validly make them more subjectively probable".


Group orbit

When we have a [3gd group] [3t9 acting] on a set, we are often interested in how the group acts on a particular [group_element element]. One natural way to do this is to look at the set of all the elements that different group actions will take the starting element to. This is called the orbit.

#Definition
Let $X$ be a set, with element $x \in X$, and let $G$ be a group acting on $X$. Then the orbit of $x$ is $Gx = \{gx : g \in G\}$.

#Properties
The set $X$ is partitioned by group orbits - each element defines its own orbit, and two orbits containing the same element must be the same, because every action has an inverse action. This gives the fact that [4j5 cosets partition a group] as the special case where $X$ is a subgroup of $G$.
#See Also

 - [4l8 The orbit-stabiliser theorem].


Subjective probability

[summary:  What does it *mean* to say that a fair coin has a 50% probability of landing heads?

Imagine flipping a coin and slapping it against your wrist. It's already landed either heads or tails. The fact that you don't know whether it's heads or tails is a fact about _you,_ not a fact about the coin. Ignorance is in the mind, not in the world.

Subjective probabilities are a tool for quantifying uncertainty about the world. In your mind's representation of the coin, you're unsure about which way the coin came up. Those probabilities, represented in your brain, are your subjective probabilities.

[1bv] is concerned with the formalization, study, and manipulation of subjective probabilities.

Several [7ry coherence theorems] suggest that classical probability is a *uniquely* good way to quantify subjective uncertainty.  This in turn means that while probabilities may be 'subjective' in the sense of existing in our minds rather than outside reality, the laws governing the manipulation and [1ly updating] of these probabilities are solid and determined.]

What does it *mean* to say that a flipped coin has a 50% probability of landing heads?

There are multiple ways to answer this question, depending on what you mean by "probability". This page discusses "subjective probabilities," which are a tool for quantifying your personal uncertainty about the world.

Imagine flipping a coin and slapping it against your wrist. It's already landed either heads or tails. The fact that you don't know whether it's heads or tails is a fact about _you,_ not a fact about the coin. Ignorance is in the mind, not in the world.

So your mind is representing the coin, and you're unsure about which way the coin came up. Those probabilities, represented in your brain, are your subjective probabilities. [1bv] is concerned with the formalization, study, and manipulation of subjective probabilities.

If probabilities are simply subjective mental states, what does it mean to say that probabilities are "good," "correct," "accurate," or "true"? The subjectivist answer, roughly, is that a probability distribution becomes more accurate as it puts more of its probability mass on the true possibility within the set of all possibilities it considers.  For more on this see [4yj].

Subjective probabilities, given even a small [ grain of truth], will become *more* accurate as they interact with reality and execute [1ly Bayesian updates].  Your subjective belief about "Is it cloudy today?" is materially represented in your brain, and becomes more accurate as you look up at the sky and causally interact with it: light from the clouds in the sky comes down, enters your retina, is transduced to nerve impulses, processed by your visual cortex, and then your subjective belief about whether it's cloudy becomes more accurate.

'Subjective probability' designates our view of probability as an epistemic state, something inherently in the mind, since reality itself is not uncertain.  It doesn't mean 'arbitrary probability' or 'probability that somebody just made up with no connection to reality'.  Your belief that it's cloudy outside (or sunny) is a belief, but not an arbitrary or made-up belief.  The same can be true about your statement that you think it's 90% likely to be sunny outside, because it was sunny this morning and it's summer, even though you're currently in an interior room and you haven't checked the weather.  The outdoors itself is not wavering between sunny and cloudy; but your *guess* that it's 9 times more likely to be sunny than cloudy is not ungrounded.

Several [7ry coherence theorems] suggest that classical probabilities are a *uniquely* good way of quantifying the relative credibility we attach to our guesses; e.g. that even if it's probably sunny, it's still *more* likely for it to be cloudy outside than for the Moon to be made of green cheese.  This in turn says that while the probabilities themselves may exist in our minds, the laws that govern the manipulation and updating of these probabilities are as solid as any other mathematical fact.

For an example of a solid law governing subjective probability, see Arbital's [1zq guide to Bayes' rule].


Lagrange theorem on subgroup size: Intuitive version

[summary: Lagrange's Theorem gives us a restriction on the size of a subgroup of a finite group: namely, subgroups have size dividing the parent group.]

Given a finite [-3gd] $G$, it may have many [576 subgroups].
So far, we know almost nothing about those subgroups; it would be great if we had some way of restricting them.

An example of such a restriction, which we do already know, is that a subgroup $H$ of $G$ has to have [3gg size] less than or equal to the size of $G$ itself.
This is because $H$ is contained in $G$, and if the set $X$ is contained in the set $Y$ then the size of $X$ is less than or equal to the size of $Y$.
(This would have to be true for any reasonable definition of "size"; the [4w5 usual definition] certainly has this property.)

Lagrange's Theorem gives us a much more powerful restriction: not only is the size $|H|$ of $H$ less than or equal to $|G|$, but in fact $|H|$ divides $|G|$.

%%hidden(Example: subgroups of the cyclic group on six elements):
*A priori*, all we know about the subgroups of the [-47y] $C_6$ of order $6$ is that they are of order $1, 2, 3, 4, 5$ or $6$.

Lagrange's Theorem tells us that they can only be of order $1, 2, 3$ or $6$: there are no subgroups of order $4$ or $5$.
Lagrange tells us nothing about whether there *are* subgroups of size $1,2,3$ or $6$: only that if we are given a subgroup, then it is of one of those sizes.

In fact, as an aside, there are indeed subgroups of sizes $1,2,3,6$:

- the subgroup containing only the identity is of order $1$
- the "improper" subgroup $C_6$ is of order $6$
- subgroups of size $2$ and $3$ are guaranteed by [4l6 Cauchy's theorem].

%%

# Proof

In order to show that $|H|$ divides $|G|$, we would be done if we could divide the elements of $G$ up into separate buckets of size $|H|$.

There is a fairly obvious place to start: we already have one bucket of size $|H|$, namely $H$ itself (which consists of some elements of $G$).
Can we perhaps use this to create more buckets of size $|H|$?

For motivation: if we think of $H$ as being a collection of symmetries (which we can do, by [49b Cayley's Theorem] which states that all groups may be viewed as collections of symmetries), then we can create more symmetries by "tacking on elements of $G$".

Formally, let $g$ be an element of $G$, and consider $gH = \{ g h : h \in H \}$.

Exercise: every element of $G$ does have one of these buckets $gH$ in which it lies.
%%hidden(Show solution):
The element $g$ of $G$ is contained in the bucket $gH$, because the identity $e$ is contained in $H$ and so $ge$ is in $gH$; but $ge = g$.
%%

Exercise: $gH$ is a set of size $|H|$. %%note:More formally put, [-4j8].%%
%%hidden(Show solution):
In order to show that $gH$ has size $|H|$, it is enough to match up the elements of $gH$ [499 bijectively] with the elements of $|H|$.

We can do this with the [-3jy] $H \to gH$ taking $h \in H$ and producing $gh$.
This has an [4sn inverse]: the function $gH \to H$ which is given by pre-multiplying by $g^{-1}$, so that $gx \mapsto g^{-1} g x = x$.
%%

Now, are all these buckets separate? Do any of them overlap?

Exercise: if $x \in rH$ and $x \in sH$ then $rH = sH$. That is, if any two buckets intersect then they are the same bucket. %%note:More formally put, [4j5].%%
%%hidden(Show solution):
Suppose $x \in rH$ and $x \in sH$.

Then $x = r h_1$ and $x = s h_2$, some $h_1, h_2 \in H$.

That is, $r h_1 = s h_2$, so $s^{-1} r h_1 = h_2$.
So $s^{-1} r = h_2 h_1^{-1}$, so $s^{-1} r$ is in $H$ by closure of $H$.

By taking inverses, $r^{-1} s$ is in $H$.

But that means $\{ s h : h \in H \}$ and $\{ r h : h \in H\}$ are equal.
Indeed, we show that each is contained in the other.

- if $a$ is in the right-hand side, then $a = rh$ for some $h$. Then $s^{-1} a = s^{-1} r h$; but $s^{-1} r$ is in $H$, so $s^{-1} r h$ is in $H$, and so $s^{-1} a$ is in $H$.
Therefore $a \in s H$, so $a$ is in the left-hand side.
- if $a$ is in the left-hand side, then $a = sh$ for some $h$. Then $r^{-1} a = r^{-1} s h$; but $r^{-1} s$ is in $H$, so $r^{-1} s h$ is in $H$, and so $r^{-1} a$ is in $H$.
Therefore $a \in rH$, so $a$ is in the right-hand side.
%%

We have shown that the "[4j4 cosets]" $gH$ are all completely disjoint and are all the same size, and that every element lies in a bucket; this completes the proof.


Proportion

A **proportion** is a representation of one value as a [-fraction] or [-multiple] of another value. It is used to provide a sense of relative magnitude.

If some value $a$ is proportional to another value $b$, it means that $a$ can be expressed as $c \times b$, where $c$ is some constant, also known as the *constant of proportionality* (or sometimes just the *proportion*) of $a$ to $b$.

For example, no matter how large a domino is, it is always twice as long as it is wide. Then, the proportion of the length compared to the width can be said to be $2$, and we can write $l = 2w$ where $w$ is the width and $l$ is the length.

## Notation

To write that $a$ is proportional to $b$, the notation $a \propto b$ is used. For example, the surface area of an object of a specific shape is proportional to the square of its length, so we write $A \propto L^2$ where $A$ is area and $L$ is length.

## Percentages

Percentages are a way of describing proportions relative to the number 100 to make them easier to understand. When people refer to a "percentage basis" for calculating something, they are specifically talking about this type of proportion.

To get a percentage from a proportion, simply multiply it by 100. In our domino example, we multiply $2$ by $100$ to find that the length is equal to $200\%$ of the width.

Percentages only work with dimensionless proportions — that is, where the constant of proportionality has no [unit units]. For example, if a car requires $6.7$ litres of gas to travel $100$ kilometres, it makes no sense to say that the fuel efficiency of the car (which is a constant of proportionality for fuel used versus distance travelled) is $6.7\%$ litres per kilometre. However, when comparing the fuel efficiencies of two cars, if one car has an efficiency of $5.2$ L/100km and another has one of $6.8$ L/100km, it makes sense to say that the first car is $131\%$ as efficient as the second.

## Common examples of proportions

The [-circumference] of a [-circle] is proportional to its [-diameter]. We write $C \propto d$, and the constant of proportionality is $\pi = 3.14159265\ldots$.

The [-area] of a circle is proportional to the square of its radius. We write $A \propto r^2$, and the constant of proportionality is also $\pi$.

The rate of growth of an [-4ts] function is proportional to the current value of the function. We write $\frac{df}{dt} \propto f$, or $\Delta f \propto f$ if considering discrete values.

The gravitational potential energy of an object in a constant gravitational field is proportional to its height relative to some zero point. We write $E \propto h$, and the constant of proportionality is the weight of the object, its mass multiplied by the gravitational acceleration in the field.

The amount of current flowing through a specific wire is proportional to the potential difference, or voltage, between the endpoints of the wire. We write $I \propto V$, and the constant of proportionality is the resistance of that wire.

In an ideal gas, the pressure of the gas is proportional to its temperature, if the volume and number of particles in the gas is held constant. This is one property of the ideal gas law in physics. We write $P \propto T$.

[todo: proportions as ratios]


Cardinality

[summary(brief): The **cardinality** of a [-3jz] is a formalization of the "number of [5xy elements]" in the set.]

[summary: If $A$ is a finite set then the cardinality of $A$, denoted $|A|$, is the number of elements $A$ contains. When $|A| = n$, we say that $A$ is a set of cardinality $n$. There exists a [499 bijection] from any finite set of cardinality $n$ to the set $\{0, ..., (n-1)\}$ containing the first $n$ natural numbers.

Two infinite sets have the same cardinality if there exists a bijection between them. Any set in bijective correspondence with [45h $\mathbb N$] is called __countably infinite__, while any infinite set that is not in bijective correspondence with $\mathbb N$ is call **[2w0 uncountably infinite]**. All countably infinite sets have the same cardinality, whereas there are multiple distinct uncountably infinite cardinalities.]

[summary(technical): The cardinality (or size) $|X|$ of a set $X$ is the number of elements in $X.$ For example, letting $X = \{a, b, c, d\}, |X|=4.$

[todo: technical summary of infinite cardinality]]

The **cardinality** of a [-3jz] is a formalization of the "number of elements" in the set.

Set cardinality is an [-53y]. Two sets have the same cardinality if (and only if) there exists a [499 bijection] between them.

## Definition of equivalence classes

### Finite sets

A set $S$ has a cardinality of a [-45h] $n$ if there exists a bijection between $S$ and the set of natural numbers from $1$ to $n$. For example, the set $\{9, 15, 12, 20\}$ has a bijection with $\{1, 2, 3, 4\}$, which is simply mapping the $m$th element in the first set to $m$; therefore it has a cardinality of $4$.

We can see that this equivalence class is [ well-defined] — if there exist two sets $S$ and $T$, and there exist bijective functions $f : S \to \{1, 2, 3, \ldots, n\}$ and $g : \{1, 2, 3, \ldots, n\} \to T$, then $g \circ f$ is a bijection between $S$ and $T$, and so the two sets also have the same cardinality as each other, which is $n$.

The cardinality of a finite set is always a natural number, never a fraction or decimal.

### Infinite sets

Assuming the [69b axiom of choice], the cardinalities of infinite sets are represented by the [aleph_numbers]. A set has a cardinality of $\aleph_0$ if there exists a bijection between that set and the set of *all* natural numbers. This particular class of sets is also called the class of [-countably_infinite_sets].

Larger infinities (which are [2w0 uncountable]) are represented by higher Aleph numbers, which are $\aleph_1, \aleph_2, \aleph_3,$ and so on through the [ordinal ordinals].

**In the absence of the Axiom of Choice**

Without the axiom of choice, not every set may be [55r well-ordered], so not every set bijects with an [-ordinal], and so not every set bijects with an aleph.
Instead, we may use the rather cunning [Scott_trick].

%%todo: Examples and exercises (possibly as lenses) %%

%%todo: Split off a more accessible cardinality page that explains the difference between finite, countably infinite, and uncountably infinite cardinalities without mentioning alephs, ordinals, or the axiom of choice.%%


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