Showing that the alternating group of degree n is normal

Question

For each natural number $n$, let $V_n$ be the subset of the symmetric group $S_n$ defined by $$V_n = \{(i j)(k l) | i,j,k,l \in \{1,\ldots, n\}, i \neq j, \text{ and } k \neq l\},$$ that is, $V_n$ is the set of all products of two 2-cycles. Let $A_n$ be the subgroup of $S_n$ generated by $V_n$; the group $A_n$ is called the alternating group of degree $n$. For any $\sigma\in S_n$ define the set $$\sigma A_n\sigma^{-1}=\{\sigma \tau \sigma^{-1} \mid \sigma\in S_n, a\in A_n\}.$$ Prove that $\sigma A_n\sigma^{-1}=A_n$ for any $\sigma\in S_n$ (that is, $A_n$ is a normal subgroup of $S_n$).

I'm thinking that maybe I can show that conjugation preserves cycle type so that if $s \in A_n$ then $\sigma s \sigma^{-1} \in A_n$, but I am not sure if this is the correct argument, or if there is a better argument.

Answer 1

An alternative approach to the cycle type argument is to show that $[S_n : A_n] = 2$, which implies that $A_n$ is normal.

The fact that conjugation preserves the length of a cycle will give you the result.

If $\mu$ is the $m$ cycle $(x_1 x_2 \dots x_m)$, then \begin{align*} \tau \mu \tau^{-1} = (\tau(x_1) \tau(x_2) \dots \tau(x_m)) \end{align*} since $\tau\mu\tau^{-1}(\tau(x_i)) = \tau(\mu(x_i)) = \tau(x_{i+1})$. So $\tau \mu \tau^{-1}$ is again an $m$ cycle.

Therefore, conjugation by $\tau \in S_n$ preserves the length of $\mu$. Now, recall that if $n \geq 3$, ($n\leq 2$ implies $S_n$ is abelian) $A_n$ is generated by the $3$-cycles. Hence, if $\sigma \in A_n$ then $\sigma = \mu_1\dots \mu_k$ where $\mu_i$ are $3$-cycles. Then \begin{align*} \tau \sigma \tau^{-1} & = \tau(\mu_1\dots \mu_k)\tau^{-1} \\ & = \tau \mu_1 \tau^{-1} \dots \tau\mu_k\tau^{-1} \end{align*} which is a product of $3$-cycles.

(Note that the $[S_n:A_n]=2$ proof is much easier).

Answer 2

Since multiplying by $(12)$ maps $A_n$ to $S_n \setminus A_n$ and vice versa, we see that $|A_n|=|S_n \setminus A_n|$, so $|S_n|=2|A_n|$ and $[S_n:A_n]=2$. But subgroups of index 2 are always normal.

Just note that if $[G:H]=2$ then the left cosets of $H$ are $H$ and $G\setminus H$, and those are also the right cosets for the same reason.

Answer 3

Consider Homomorphism $\phi:S_n\to$ <{+1,-1},.>
WIth $\phi(a)=sign(a)$
As Alternating group by defination permutation with sign 1
It is easy to see $A_n$ is kernal of above map.
Every kernel of homomorphism is normal subgroup.