[[Symmetric group]]
# Alternating group

The **alternating group** $\mathrm{Alt}_{n}$ of degree $n$ is the [[Kernel of a group homomorphism|kernel]] of the [[alternating character]] $\sgn : \mathrm{S}_{n} \to \mathrm{S}_{2}$, #m/def/group 
and therefore a [[normal subgroup]] made up of all even permutations.
For $n \geq 2$ we have the [[Group extension#^split]] (and hence [[semidirect product]])
$$
\begin{align*}
1 \to \mathrm{Alt}_{n} \hookrightarrow \mathrm{S}_{n}  \twoheadrightarrow 2 \to 1
\end{align*}
$$

## Simplicity

An important property of the alternating group $\mathrm{Alt}_{n}$ for $n \geq 5$ is that it is a [[simple group]]. #m/thm/group 
This is proven using the following lemmata

1. $\mathrm{Alt}_{n}$ for $n \geq 3$ is generated by 3-cycles. ^S1
2. If $N \trianglelefteq \mathrm{Alt}_{n}$ with $n \geq 3$ contains a 3-cycle, then $N = \mathrm{Alt}_{n}$. ^S2
3. Every nontrivial $N \trianglelefteq\mathrm{Alt}_{n}$ for $n \geq 5$ contains a 3-cycle. ^S3

> [!check]- Proof
> Since pairs of transpositions generate $\mathrm{Alt}_{n}$ for $n \geq 3$ by construction,
> we need only show that any pair of transpositions can be written as a product of 3-cycles.
> Noting that $(a,b)=(b,a)$,
> the following list exhausts any pair of transpositions:
> $$
> \begin{align*}
> (a,b)(a,b) &= e \\
> (a,b)(c,d) &= (a,c,b)(a,c,d) \\
> (a,b)(a,c) &= (a,c,b)
> \end{align*}
> $$
> proving [[#^S1]].
> 
> Now $\mathrm{Alt}_{n}$ for $n \geq 3$ can in fact be generated only from 3-cycles of the form $(i,j,k)$ with $i,j$ fixed in $\mathbb{N}_{n}$,
> since every 3-cycle can be expressed as such:
> $$
> \begin{align*}
> (i,a,j) &= (i,j,a)^2 \\
> (i,a,b) &= (i,j,b)(i,a,j) \\
> (j,a,b) &= (i,b,j)(i,j,a) \\
> (a,b,c) &= (j,c,a)(j,a,b)
> \end{align*}
> $$
> Now assume $N \trianglelefteq \mathrm{Alt}_{n}$ contains a 3-cycle, say $(i,j,a)$.
> By normality it follows for any $b \in \mathbb{N}_{n} \setminus \{ i,j,a \}$
> $$
> \begin{align*}
> (i,j,b) = [(i,j)(a,b)] (i,j,a)^2 [(i,j)(a,k)]^{-1} \in N
> \end{align*}
> $$
> Hence $N$ contains all 3-cycles of the form $(i,j,b)$ and hence all 3-cycles,
> thus by [[#^S1]] it is $\mathrm{Alt}_{n}$, proving [[#^S2]].
> 
> Now let $N \trianglelefteq\mathrm{Alt}_{n}$ for $n \geq 5$ be nontrivial.
> Then one of the following holds:
> 
> **Case 1:** Suppose there exists $\sigma \in N$ with a cycle of length $\geq4$,
> so without loss of generality (by relabelling) $\sigma = (123 \dots r)\tau$
> where $(123 \dots r)$ and $\tau$ are disjoint.
> Since $(123) \in \mathrm{Alt}_{n}$, it follows
> $$
> \begin{align*}
>  \sigma^{-1}(123)\sigma (123)^{-1} &= \tau^{-1}(123\dots r)^{-1} (123)(123\dots r) \tau (123)^{-1} \\
>  &= (r \dots 321)(231\dots r) = (13r) \in N
> \end{align*}
> $$
> so $N$ contains a 3-cycle.
> 
> 
> **Case 2a:** Suppose there exists a $\sigma \in N$ which contains two disjoint 3-cycles (and nothing longer).
> Without loss of generality, $\sigma = (123)(456)\tau$ for disjoint $(123)$, $(456)$, and $\tau$.
> Since $(124) \in \mathrm{Alt}_{n}$, it follows
> $$
> \begin{align*}
> \sigma^{-1}(124)\sigma(124)^{-1} 
> &= \tau^{-1}(456)^{-1}(123)^{-1}(124)(123)(456)\tau(124)^{-1} \\
> &=(654)(321)(123)(123)(456)(421) = (14263) \in N
> \end{align*}
> $$
> which falls under case 1.
> 
> **Case 2b:** Suppose there exists a $\sigma \in N$ containing exactly one 3-cycle and otherwise only transpositions.
> Without loss of generality $\sigma = (123) \tau$ where $(123)$ and $\tau$ are disjoint,
> and $\tau^{-1}=\tau$.
> Then $\sigma^2 = (123)\tau(123)\tau = (123)^2 = (132) \in N$,
> so $N$ contains a 3-cycle.
> 
> **Case 2c:** If $N$ contains a 3-cycle we are already done.
> 
> **Case 3:** The only remaining possibility is that there exists a $\sigma \in N$ which is a product of disjoint transpositions,
> and an even number thereof since $N \trianglelefteq \mathrm{Alt}_{n}$.
> Without loss of generality $\sigma = (12)(34)\tau$ with $(12)$, $(34)$, $\tau$ disjoint and $\tau^{-1} = \tau$.
> Then
> $$
> \begin{align*}
> \sigma^{-1}(123)\sigma(123)^{-1} &= \tau^{-1}(34)(12)(123)(12)(34)\tau(132) \\
> &= (34)(12)(123)(12)(34)(132) = (13)(24) \in N
> \end{align*}
> $$
> whence
> $$
> \begin{align*}
> (13)(24)(135)(13)(24)(135)^{-1} = (135) \in N
> \end{align*}
> $$
> thus $N$ contains a 3-cycle.
> 
> This proves [[#^S3]] and therewith the simplicity of $\mathrm{Alt}_{n}$ for $n \geq 5$.
> <span class="QED"/>

Note that $\mathrm{Alt}_{2}$ is trivial, $\mathrm{Alt}_{3} \cong \mathbb{Z}_{3}$ is Abelian and simple,
but $\mathrm{Alt}_{4}$ is not simple as $\{ e,(12)(34),(13)(24),(14)(23) \} \triangleleft \mathrm{Alt}_{4}$.
See [[Decomposition of S4]].

## Properties

1. $\mathrm{Alt}_{n}$ is $(n-2)$-[[Multiply transitive permutation group|transitive]].

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