Show that if $SS=S $ then $S$ is a subgroup 
Let $S$ be a non empty finite subset of a group $G$ such that $SS=S $ then $S$ is a subgroup

First of all I want to show that $1 \in S $ my idea is to prove it by the absurd. So let's suppose $1 \notin S $. Now since $S $ is non empty,  take $s \in S =SS$. Therefore $s=s_1 s_2 $ with $s_1,s_2$ different from $s$ (if not it is easy to see that $1\in S $) so the idea behind this reasoning is to show that the only way for this to be possible is that $S $ is Infinite. However, when writing for example $s_1$ as a product of two elements of $S $ there is no reason why one of those cant be $s $ for example. And that detail is getting me stuck. Am I on the right track? Any other hint or possibility?
 A: Write $S=\{s_1,s_2,...,s_n\}$ with $n\geq 1$. Since $SS=S$, we have
$$s_1s_1=s_{k_1}, s_1s_2=s_{k_2}, s_1s_3=s_{k_3}, ..., s_1s_n=s_{k_n}$$
where $k_1,...,k_n\in\{1,2,..,n\}$. The $k_i$ must be distinct otherwise we have
$$s_1s_i=s_1s_j$$
for some $i\not=j$ hence $s_i=s_j$ which is impossible. Then we have 
$$\{k_1,k_2,...,k_n\}=\{1,2,...,n\}$$
so there is a $k_i$ such that $k_i=1$ hence
$$s_1s_i=s_1$$ 
which gives $s_i=1$ so that $1\in S$.
A: Your approach does not look likely to work. For instance, consider the group $\mathbb Z/3\mathbb Z$ and the subset $S=\{\bar 1,\bar 2\}$. Then:
$$\bar 2 = \bar 1+\bar 1$$
$$\bar 1 = \bar 2 + \bar 2.$$
In this instance, it is true that every element can be written as a product of elements distinct from it, yet $S$ is not a subset and $SS\neq S$. Therefore, you cannot finish the argument just by using that $s=s_1s_2$ for $s_1,s_2\in S\setminus \{s\}$ for any $s\in S$.
A more promising idea is to drop the argument by contradiction and consider, for any $s\in S$, the set $\{s,s^2,s^3,\ldots\}$. You can prove first that this is a subset of $S$ and therefore finite. Then, you can find within this set both the identity, and the inverse of $s$.
