Showing a set is closed under inverses Let $G$ be a finite group, say a group with $n$ elements, and let $\emptyset \neq S \subseteq G$. Suppose $e \in S$, and that $S$ is closed with respect to multiplication. Prove
that $S$ is a subgroup of $G$.

Let $G = \{a_1, \ldots, a_n\}.$ Let $a_i \in G.$ Then any number of $a_ia_1, a_ia_2, \ldots, a_ia_n \in S,$ since $S$ is closed under multiplication. One of those $a_ia_j$ is $e$ since $e \in S.$ Then either $a_i$ or $a_j$ is $a^{-1}.$ Would this argument work?
 A: You do not really need $G$ to be finite...
It is enough if you assume $S$ is finite...
let $S=\{a_1,a_2,a_3,\cdots,a_n\}$ and set $e=a_1$ identity element...
Take $a_2\in S$ and consider $M=\{a_2a_j : 1\leq j\leq n\}$..
Check 1 : Prove that $a_2a_j\neq a_2a_i$ for any $i\neq j$..
Chek 2 : So cardinality of $M$ is ???
Check 3 : As $S$ is closed under multiplication $M\subseteq S$
Check 4 : Using cardinality of $M$ see that $M=S$ (this is where you use finiteness of $S$)
Check 5 : As $e\in S$ we should have $e\in M=\{a_2a_j : 1\leq j\leq n\}$.. So, $e=a_2a_j$ for some $j$... 
I have already said more than enough....  :D 
Note : For $S$ to be invertible you want inverse of $a_2$ to be in $S$.. 
A: This is essentially the fact that a finite monoid with cancellation must be a group. It's the group analogue of the fact that a finite domain is a field.
In both cases, the proof comes from considering the map $x \mapsto ax$. Cancellation implies that this map is injective. Finiteness then implies that the map is surjective. So $1$ is in the image of the map and this gives an inverse for $a$.
Cancellation holds because your monoid is inside a group. 
