Finite Group Proving finite order of elements and Subgroup Question The question is as follows 
Let G be a finite group.
(i) Prove that every element of G has finite order.
For this want to use the idea that if G is finite then for a in G, $a^{n}$ = $e$ for some n in $Z^{+}$
and treat a is a generator of a cyclic group that is a subset of G the the order of each $<a_{i}>$   must be less than or equal to G. since G is of finite order each $<a_{i}>$ is also of finite order. but i am really not sure i can treat this group in such a way?
(Secondly) Suppose that H is a subset of G which is non-empty and, for any
a, b in H, ab is also in H. Prove that H is a subgroup of G.
For this i want to try and use that since ab is in H $(ab)^{n}$ should be in H and for some n in $Z^{+}$  
$(ab)^{n}$ = e but im not sure my assumption in part 1 is correct... could someone point me in the right direction please?
EDIT
I think i understand since H is finite ( cause G is finite) I understand why for any h in H
$<h>$ must be a subset of H but how do we prove that $(h)^{2}$ is in H oh! Perhaps we use for any a,b in H ab is H and just pick h twice so hh must be in H thus $(h)^{n}$ must be in H and $(h)^n$ = e for some n in $Z^{+}$ by G and this must be true for all h in H ?
EDIT (ii)
Ok let a and b be in H and left $a^{l} = e$ and  $b^{k}= e$ for some l and k in $Z^{+}$
then $b^{k}$$a^{l}$ is in H and $b^{k}$$a^{l}$ = e so 
$b^{-k}$$b^{k}$$a^{l}$ = $b^{-k}$
so $a^{l}$ = $b^{-k}$ 
Can we assume from this that every h in H has an inverse because every $h^{n}$ has one?
Im still curious about above but i think that anon actually had answered my question of where the i could find the Inverse element from at the end of his post. ( i just need to digest the information a bit)
Thanks so much to all of you, its very helpful to be able ask a question and get such a wonderful number of different ways of looking at the problem.
 A: For (i) you are right. Take the cyclic group $\langle g\rangle\subseteq G$ generated by an arbitrary $g\in G$; since it is a subset of a finite set ($G$ is finite) it must be finite, so the order of $g$ is finite (note that an equivalent way of defining the order of $g$ is as $\#\langle g\rangle$, if you're not already using this definition).
For (ii), since $a,b\in G\implies ab\in G$, there are two other group properties you need to check: first, that $H$ contains the identity $e\in G$, and second, that for all $a\in H$, the inverse $a^{-1}\in H$ is too. For this part, finiteness of $G$ is critical (it is not generally true that a subset closed under the group operation need be a subgroup of an infinite group; take the nonnegative naturals under addition as a subset of the group of integers under addition).
To do this, prove that for any $h\in H$, we have $\langle h\rangle\subseteq H$. Then show both $e$ and $h^{-1}$ are in $\langle h\rangle$, and hence in $H$ as well. This would establish all three properties of $H$'s being a subgroup.
A: Hint for 1: Suppose some element $g\in G$ had infinite order. That is, there is no $n$ such that $g^n = e$. I claim each $g^i$ ($i\in\Bbb{N}$) is distinct. (Suppose $g^i = g^j$ for $i,j\in\Bbb{N}$ and obtain a contradiction.)
For your second part, let $G = \Bbb{Z}$ and $H = \Bbb{N}$ (both under addition). $\Bbb{N}$ satisfies that for all $n,m\in\Bbb{N}$, $n + m\in\Bbb{N}$ (and certainly $\Bbb{N}\neq\emptyset$), but is $\Bbb{N}$ a subgroup of $\Bbb{Z}$?
EDIT: This gives an example of why the second statement is not true for infinite groups. Does this suggest a reason that it might be true for finite group? That is, why does it fail in this case, and why can't that happen in a finite group (the reason is similar to the reasoning for problem 1).
A: Let $a$ be an element of $G$.
Consider the set $S = \{a, a^2, \dots, a^n, \dots\}$.
Since $S$ is finite, there exist positive integers $n < m$ such that $a^n = a^m$.
Hence $a^{m-n} = 1$.
A: Let $a$ be an element of $G$.
If $a$ is of infinite order, $G$ contains a subgroup of infinite order.
Hence $G$ is not a finite group.
This is a contradiction.
