Prove the following about $S_n$ and $A_n$ 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? 
 A: 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).
A: 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:


*

*$\sigma\tau\sigma^{-1}\tau^{-1}\in A_n$ is equivalent to $$\operatorname{sgn}(\sigma\tau\sigma^{-1}\tau^{-1}) = +1.$$

*The signum map $$\operatorname{sgn} : S_n\to\mathbb{Z}_2$$
is a group homomorphism.

*$\mathbb{Z}_2$ is abelian.

A: 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 $+1$.
A: 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.
A: Given a permutation $\sigma$, it can be expressed as a product of transpositions in many ways, but the parity (even or odd) of the number of transpositions would be the same for all these expressions.   In fact, the parity depends only on the length of the disjoint cycles in $\sigma$ and not on the permutation itself.  These lengths are the same for $\sigma^{-1}$ as for $\sigma$, whence $\sigma$ and $\sigma^{-1}$ have the same parity.  
Also, when permutations are composed (as in $\sigma \tau \sigma^{-1} \tau^{-1}$), the parity of the product is the sum of the individual parities. In your case the composition is $\sigma \tau \sigma^{-1} \tau^{-1}$, and its parity is the sum of twice the parity of $\sigma$ and twice the parity of $\tau$, and hence must be even (regardless of the parity of $\sigma$ or of $\tau$). 
