Reason(s) Why the Set of Odd Permutations is Not a Subgroup of $S_n$ 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!
 A: 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.
A: You are taking some intuitions from arithmetic and using them incorrectly for the arithmetic of permutations.
The identity permutation is the product of $0$ transpositions. Since $0$ is an even number, the identity is an even permutation. If you are uncomfortable with the product of zero permutations, write the identity as the product of two transpositions: $(1,2)(1,2)$.
Yes, the product of two odd numbers is odd. But if you multiply the product of $m$ transpositions by the product of $n$ you have multiplied $m+n$ transpositions together. The sum of two odd numbers is even.
A: The set of odd permutations does not contain a neutral element. This can only be the identical permutation, and this permutation is even.
Every subgroup is a group and every group contains a neutral element.
The other argument : If we have two odd permutations, the number of transpositions of the product is modulo $2$ equal to the sum of the numbers of transpositions of the permutations. If we have two odd permutations, the product is therefore even because the sum of two odd numbers is even.
A: You are correct about odd*odd=even. But this not happens in our case
Regarding your question, all of the answers provided by other authors are absolutely great. I just want to provide another point of view. 
I presume that you have learn the $sign$ of a permutation. So we have $sgn (\sigma) = 1$ if a permutation is $even$ and $sgn (\sigma) = -1$ if a permutation is $odd$. 
So if you have two odd permutations and consider $\sigma \times\sigma$, the sign of $\sigma \times\sigma$ would be $-1*-1 = 1$. Therefore $\sigma \times\sigma$ will be an even permutation.
Also for the identity. It must be even, otherwise, the sign of $id*\sigma$ will change.
A: I think you get the answer. But i would say that its possible to show that if $\space H\lt S_n$ then $H\cap{A_n}\lt An$, "H has only even permutations", or $H\cap{A_n}$ has half even permutations and half odd. So its NOT possible to have a subgroup of $S_n$ which has only odd permutations. ($A_n$ is the subgroup of all even permutations).
I hope this can help you. 
