1. ## Permutation subgroup

Prove the set of even permutation in $\displaystyle S_{n}$ forms a subgroup of $\displaystyle S_{n}$.

How should I start this?

Prove the set of even permutation in $\displaystyle S_{n}$ forms a subgroup of $\displaystyle S_{n}$.

How should I start this?
This is referred to an "alternating group" $\displaystyle A_n$.

Let $\displaystyle G$ be a group and $\displaystyle H$ a non-trivial subset of $\displaystyle G$. What makes $\displaystyle H$ a subgroup? What conditions must be met?

Now do the same thing here. Show all of these conditions are satisfied.

Suppose that $\displaystyle H \subset S_{n}$ with $\displaystyle H= \{ a_{1},a_{2}, ... , a_{n} \}$ 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 $\displaystyle \alpha , \beta \in H$.

Then $\displaystyle \alpha = (a_{1} a_{2})(a_{3}a_{4})$ and $\displaystyle \beta = (b_{1}b_{2})(b_{3}b_{4})$

Now, $\displaystyle \alpha \beta$ 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 $\displaystyle \alpha$ is its own inverse, thus satisfied the conditions for H to be a subgroup of Sn.

Am I doing something right here?

Thanks.

$\displaystyle \alpha$ is even and $\displaystyle \beta$ is even then $\displaystyle \alpha \beta$ is even. You mentioned that but you did not show it. Why is that? Remember you need to show this to show that $\displaystyle H$ is closed.