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?

To show $H$ is a subgroup you must show it's closed, contains the identity, and contains inverses. But if it's closed, non-empty, and contains inverses, then it's guaranteed to contain the identity, because it's guaranteed to contain something, say, $x$, then $x^{-1}$, then $xx^{-1}$, which is the identity. 
$H$ is assumed closed, so if it contains $a$ and $b$, it contains $ab$. But $a$ and $b$ don't have to be different: if it contains $a$, it contains $a$ and $a$, so it contains $aa$, which is $a^2$. But then it contains $a$ and $a^2$ so it contains $aa^2$ which is $a^3$. Etc. 
So it contains $a,a^2,a^3,a^4,\dots$. $H$ is finite, so these can't all be different, so some pair is equal, that is, $a^i=a^j$ for some $i\ne j$. 
As for your last question, do you know any example of a group with a non-cyclic subgroup?
