Okay, let me think more about this...
Suppose that with with a being permutations that can be expressed as even number of 2-cycles, in other words, H contains all of the even permutations in Sn.
Now pick .
Then and
Now, is in H because it is only another expression of even permutations. (Do I need to justify this? Or do we already know that? I know that by the problems that I have done before)
Then is its own inverse, thus satisfied the conditions for H to be a subgroup of Sn.
Am I doing something right here?
Thanks.