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.