The Pigeon Hole Principle and the Finite Subgroup Test I am currently reading this document and am stuck on Theorem 3.3 on page 11: 

Let $H$ be a nonempty finite subset of a group
    $G$. Then $H$ is a subgroup of $G$ if $H$ is closed
    under the operation of $G$.

I have the following questions:
1.

It suffices to show that $H$ contains inverses.

I don't understand why that alone is sufficient.
2.

Choose any $a$ in $G$...then consider the sequence $a,a^2,..$ This sequence is contained in $H$ by the closure property.

I know that if $G$ is a group, then $ab$ is in $G$ for all $a$ and $b$ in $G$.But, I don't understand why the sequence has to be contained in $H$ by the closure property.
3.

By the Pigeonhole Principle, since $H$ is finite, there are distinct $i,j$ such that $a^i=a^j$.

I understand the Pigeonhole Principle (as explained on page 2) and why $H$ is finite, but I don't understand how the Pigeonhole Principle was applied to arrive at $a^i=a^j$.
4.
 Reading the proof, it appears to me that $H$ = $\left \langle a \right \rangle$ where $a\in G$.
Is this true?

I am guessing that by "closed under the operation of $G$" they mean closed under the multiplication on $G$. Then $H$ is a subgroup if it then also contains inverses because it also contains the identity. Since $H$ is finite, then so must be $\{a^k:k\in\mathbb{Z}^+\}$ since it is contained in $H$.  Thus, for some $i>j$, we must have $a^i=a^j$. Thus, $a^{i-j}$ must be the identity and $a^{i-j-1}=a^{-1}$.
$H$ is not necessaily cyclic. Here we have shown that $\{a^k:k\in\mathbb{Z}^+\}$ is a cyclic subgroup, but there is no guarantee that this is all of $H$.  We may need to perform the above process on any $b\in H\setminus\{a^k:k\in\mathbb{Z}^+\}$, etc.
