Let $\sigma, \tau \in S_n$. Prove $\sigma\tau \text{ is even} \iff \sigma \text{ and }\tau$ are both even or both odd.

Question

I know that this has to be proven both ways. If $\sigma \tau$ is even then $\sigma$ and $\tau$ are both even or both odd, and if $\sigma$ and $\tau$ are both even or both odd then $\sigma \tau$ is even.

If I let $\sigma \tau$ be even, then I know it can be written as a product of an even number of transpositions. Since $\sigma$ and $\tau$ are both in $S_n$, they must be disjoint. This means that the product of numbers of transpositions of $\sigma$ and $\tau$ will equal that of $\sigma \tau$.

If $\sigma$ is even, it must be the product of an even number of transpositions which is an odd number of permutations (though I'm not sure if I can just say that or if I need to prove it). If $\sigma$ is odd, it can be written as the product of an odd number of transpositions which is an even number of permutations.

The same can be said for $\tau$.

The product of two even numbers is even, the product of two odd numbers is even, but the product of an even and an odd number is odd. Therefore, for $\sigma \tau$ to be even, $\sigma$ and $\tau$ must be both even or both odd.

Does this make sense or am I missing something?

Answer 1

Here is your first mistake:

Since $\sigma$ and $\tau$ are both in $S_n$, they must be disjoint.

You're on the right track for the proof overall though.