Today in class, the professor went over some reasons why the set of odd permutations is not a subgroup of $S_n$. There are a couple reasons that I don't understand; those are:
1) The product of two odd permutations is even, so the group is not closed.
2) It does not contain the identity element (id = even).
These may be stupid questions, but why is the product of two odd permutations even? I thought odd * odd was always odd? Also, why is it that the identity element must be even? Thank you for the help!
Best Answer
In this situation, the fact that the product of two odd numbers is odd is not relevant. What is relevant is that the sum of two odd numbers is even. Let $\sigma$ and $\tau$ be odd permutations, then $\sigma$ and $\tau$ are both products of an odd number of transpositions, that is, $\sigma=t_{1}...t_{n}$ and $\tau=s_{1}...s_{m}$ where the $t_{i}$ and $s_{i}$ are transpositions and $n$ and $m$ are odd. We have $\sigma\tau=t_{1}...t_{n}s_{1}...s_{m}$ . Note that $\sigma\tau$ is a product of $n+m$ transpositions, and $n+m$ is even because $n$ and $m$ are odd. Therefore $\sigma\tau$ is an even permutation.