Show $S_n/A_n$ is commutative Let $S_n$ be the symmetric group and $A_n$ the alternating subgroup. I want to show $S_n/A_n$ is commutative.
Given that the index of $A_n$ in $S_n$ is $2,$ the quotient group consists of two elements, $S_n/A_n = \{ A_n, \sigma A_n \mid \sigma \notin A_n\}.$ We want to show, $A_n \sigma A_n = \sigma A_n.$ This is equivalent to $(\sigma A_n)^2 = A_n,$ which is the well know condition of being abelian.
I thought first in terms of cycles. An element of $A_n$ is a product of an even number of $2-$cycles. Since $\sigma \notin A_n,$ it can be written as a product of an odd number of $2$-cycles.
Any suggestion how to go from here?
Thanks.
 A: If $n=1,$ then the result follows trivially. So let $n\geq 2.$ Since order of $S_n/A_n$ is $2$ which is prime and any group of prime order is cyclic, therefore $S_n/A_n$ is cyclic and hence also abelian.
A: You know that the order of $S_n/A_n$ is two (for $n\ge 2$; it is trivial (and hence abelian) otherwise). There is only one group of order two up to isomorphism: $\Bbb Z_2$. (One way to see this is to construct a Cayley table.) But $\Bbb Z_2$ is cyclic and hence abelian.

Another proof, somewhat overkill, is to use the First Isomorphism Theorem, with
$$\begin{align}
\varphi: S_n&\to (\{1,-1\},\times),\\
\rho &\mapsto {\rm sgn}(\rho).
\end{align}$$
I will leave this as an exercise.

One could argue that the second proof above is what goes on behind the scenes of the first proof.
A: To show that / is commutative, you need to show that = for all ∉. This is equivalent to showing that ()2=, which means that  is an element of order 2 in /.
Since  is an element of  that is not in , it can be written as a product of an odd number of 2-cycles. You can write  as a product of 2-cycles in the following way:
 = (11)(22) ... ()
where  and  are elements of {1, 2, ..., }.
Since  is an odd permutation,  is odd. We can rewrite  as follows:
 = (11)(22) ... ()
Since  is the group of even permutations, and each 2-cycle is an even permutation, we can simplify the expression to:
 = (11)(22) ... ()
Since  is odd, the expression  is equal to . Therefore, =.
Now we can show that ()2=:
()2 =  =  = 2 = 
Since ()2=,  is an element of order 2 in /. This means that / is an abelian group, and therefore / is commutative.
Hope this helps.
