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.