# Showing that the alternating group of degree n is normal

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.

• Since multiplying by $(12)$ maps $A_n$ to $S_n \setminus A_n$ and vice versa, we see that $|A_n|=|S_n \ A_n|$, so $|S_n|=2|A_n|$ and $[S_n:A_n]=2$. But subgroups of index 2 are always normal. – C Monsour Sep 12 '18 at 1:51
• I had said that using the fact that conjugation preserves cycle type is the right approach. While this is one way to prove this, it is not the easiest way. Showing that $[S_n : A_n] = 2$ is probably best. – matt stokes Sep 12 '18 at 5:32

## 3 Answers

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).

• What is the approach that involves conjugation preserving cycle types? – Wesley Sep 12 '18 at 2:31
• @Miles Davis I have updated my post. Hopefully this helps. – matt stokes Sep 12 '18 at 5:18
• Don't I need to show explicitly that $S_n$ is generated by 3-cycles (that fact isn't given in this definition of $A_n$). Also, why is it true that if $\sigma \in A_n$ then $\sigma = \mu_1 \mu_2 \dots \mu_k$? Shouldn't these $\mu$'s also have powers on them, since they are a generating set? – Wesley Sep 12 '18 at 5:46
• Yes, sorry I had to edit this a lot. The point is that $\sigma \in A_n$ iff $\sigma$ is the product of $3$-cycles. $\tau \sigma \tau^{-1}$ is again a product of $3$-cycles and so $\tau\sigma \tau^{-1} \in A_n$. Again, it is much easier to show that $[S_n : A_n] =2$ than it is to show that $A_n$ is generated by all of the $3$-cycles. – matt stokes Sep 12 '18 at 5:50
• If you except that $\sigma \in S_n$ is either even or odd (and not both), then the map $\phi : S_n \to \mathbb{Z}/2\mathbb{Z}$ given by $\sigma \mapsto 0$ if $\sigma$ is even, and $\sigma \mapsto 1$ if $\sigma$ is even, is a surjective homomorphism. Its kernel is $A_n$. Now use first isomorphism theorem to see that $[S_n : A_n]= 2$. – matt stokes Sep 12 '18 at 6:00

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.

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.