Show that the set of all even permutations in $G$ forms a subgroup of $G$. 
Show that if $G$ is any group of permutations, then the set of all even permutations in $G$ forms a subgroup of $G$.

I know that I need to show the closure, identity, and inverses properties hold. So I need to prove:
1) If $p$ and $q$ are even permutations, then so is $pq$.
2) Identity permutation is an even permutation
3) If $p$ is an even permutation, then so is $p^{-1}$
I am just unsure how to go about setting up each proof. Thanks in advance for the help.

Hint: consider the following map
$$\operatorname{sgn}(p)=\begin{cases}1,\, p\,\text{even};\\-1,\,p\,\text{odd}.\end{cases}$$
Show that it is a homomorphism from $G$ to $\{\pm1\}$. What's it's kernel?
