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.

Let $E$ be the set of even permutations in $G$ (which is presumably a group of permutations).
Let $p$ and $q$ be elements of $E$.
Check to see if $pq^{-1}$ is also an element of $E$.  (Note: this checks all three conditions simultaneously).
A permutation is called an even permutation if its expression as a product of disjoint cycles has an even number of even-length cycles.
Alternatively, a permutation is called an even permutation if it can be written as a product of an even number of transpositions.
These two definitions can be seen to be equivalent.  The second seems a bit more useful here.
So, let $p=p_1p_2\cdots p_{2k}$ and $q=q_1q_2\cdots q_{2j}$ be a representation of $p$ and $q$ as a product of transpositions.
We have that $q^{-1}=q_{2j}q_{2j-1}\cdots q_2q_1$ since transpositions are self inverses.
Thus, $pq^{-1} = p_1p_2\cdots p_{2k}q_{2j}\cdots q_2q_1$ is indeed a product of an even number of transpositions.  Furthermore, $pq^{-1}$ is an element of $G$ since $p$ and $q$ (and thus $q^{-1}$) are elements of $G$ and $G$ is closed under products and inverses.
Thus $pq^{-1}\in E$, implying that the identity is an element of $E$ (by taking $p=q$), that it is closed under inverse (by taking $p=id$), and that it is closed under products (by taking $q^{-1}$ instead of $q$) and $E$ is a subgroup of $G$.
