Show $S_n/A_n$ is commutative

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

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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.

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One could argue that the second proof above is what goes on behind the scenes of the first proof.