Prove the following about $S_n$ and $A_n$

Question

Prove that $\sigma \tau \sigma^{-1} \tau^{-1} \in A_{n}$ for any $\sigma,\tau \in S_{n}$

So I was thinking that since $A_{n}$ is a subgroup of $S_{n}$ that you can just rearrange $\sigma \tau \sigma^{-1} \tau^{-1}$ and show that it just equals two identities multiplied together which is just the identity. Then we know the identity is even which shows that it is in $A_{n}$. Does that work?

Answer 1

The permutations $\sigma$ and $\sigma^{-1}$ are both odd or both even,and the product of permutations keeps the odd/even rule; so a construct like $\sigma \tau \sigma^{-1} \tau^{-1}$ must be even (it is 4 even, 2 even + 2 odd, or 4 odd; all combinations are even).

Answer 2

You can't rearrange the product because $S_n$ is non-commutative. It is easily verified, for example, that $$(12)(23)(12)^{-1}(23)^{-1} = (12)(23)(12)(23) = (123)(123) = (132),$$ using cycle notation. I don't know how much you know about the sign of a permutation, but I would phrase the proof this way: the assignment of a sign $\pm 1$ to a permutation in $S_n$ is a homomorphism from $S_n$ to $\{\pm 1\} \cong \mathbb{Z}/(2)$. Since the latter is commutative, the product does cancel and so a commutator will have sign $+1$.

Answer 3

First note that or u can prove that $\sigma$ is even(or odd) then $\sigma^{-1}$ is also even(odd).

Case-I If $\sigma$ and $\tau$ are even(or odd) then its inverse are also even(or odd). so, we get $\sigma \tau \sigma^{-1} \tau^{-1}$ is even(if it odd then product of odd permutation must be even)

Case-II If $\sigma$ is even and $\tau$ is odd then its $\sigma \tau$ will be odd and corresponding inverses product(i.e. $\sigma^{-1} \tau^{-1}$) will be odd so we get $\sigma \tau \sigma^{-1} \tau^{-1} $ is even

Case-III follows from case-II $\sigma$ is odd and $\tau$ is even then i hope u understand.

Answer 4

As pointed out by amWhy, the rearranging you suggested does not work since $S_n$ is not abelian. However, there is still a way to make this idea work.

Hint:

1. $\sigma\tau\sigma^{-1}\tau^{-1}\in A_n$ is equivalent to $$\operatorname{sgn}(\sigma\tau\sigma^{-1}\tau^{-1}) = +1.$$
2. The signum map $$\operatorname{sgn} : S_n\to\mathbb{Z}_2$$ is a group homomorphism.
3. $\mathbb{Z}_2$ is abelian.