A problem about $SS=G?$ Suppose $G$ is a Group, $S$ is a subset of $G$. Given that $\exists a,aS\cap S\neq \emptyset, aS\cup S=G$, prove or give a counterexample: $\forall g\in G,\exists x,y\in S,s.t.xy=g\ (i.e.\; SS=G)$
I can prove the proposition is true when G is a abelian group, the prove is as follow:
wts. $\forall g'\in G, g'S^{-1}\cap S\neq \emptyset $
, here $S^{-1}=\{s^{-1}: s\in S\}$
Anti-evidence method:
if $g'S^{-1}\cap S=\emptyset $
,then $ag'S^{-1}\cap aS=\emptyset$
given that $S\cap aS\ne \emptyset$ and $S\cup aS=G$, so $ag'S^{-1}\subsetneqq S$
proposition 2: if $pS^{-1}\subset S$, then $pS^{-1}=S$
prove: Anti-evidence method: if there is $s$ in S but not in $pS^{-1}$
then $ps^{-1} \in pS^{-1}\subset S$, $p(ps^{-1})^{-1} \in pS^{-1}$
$G$ is a abelian group, so $p(ps^{-1})^{-1}=s\in pS^{-1}$, contradictory.
So the proposition is proved.
so $ag'S^{-1}\subsetneqq S$ leads to a contradiction.
then SS=G is proved.
I wonder whether the proposition holds when G is not a abelian group. Can anyone help me?(thanks)
 A: I post this question on my school's BBS, and finally got the answer. Thank you for all your help.
The proof process is as follows:

$S \cap gS\neq \emptyset,S\cup gS=G$
$\Leftrightarrow S\cap g^{-1}S\neq \emptyset,S\cup g^{-1}S=G$
therefore the "$g$" in the following proposition can be all replaced by "$g^{-1}$".

Proposition 1. If $a\notin S(a\in G)$, then $ga,g^{-1}a\in S$
Proof. Given that $S\cup gS=G$, if $a\notin S$, then $a\in gS$, $g^{-1}a\in S$; similarly $a\in g^{-1}S,ga\in S$.

Proposition 2. $S\cap Sg\neq \emptyset$
Proof by contradiction. Suppose that $S\cap Sg=\emptyset$,
$S\cap Sg=\emptyset \Rightarrow Sg\subset g^{-1}S$（deduced by $S\cup g^{-1}S=G$）
$\Leftrightarrow gSg\subset S\Rightarrow gSg\cap Sg=\emptyset$（deduced by $S\cap Sg=\emptyset$）
$\Leftrightarrow gS\cap S\notin \emptyset$, and this lead to a contradiction.

Back to the main result:
Conclusion. $SS=G$.
Proof by contradiction. suppose that there is $h\notin SS(h\in G)$, then $\forall s\in S, sg,sg^{-1}\notin S$.
Proof. $\forall s\in S$, $h=s(s^{-1}h)=(hs^{-1})s$, so $s^{-1}h,hs^{-1}\notin S$
Let $a=s^{-1}h$, using proposition 1, we can conclude that:
$a=s^{-1}h\notin S \Rightarrow gs^{-1}h\in S\Rightarrow h(gs^{-1}h)^{-1}\notin S \Leftrightarrow sg^{-1}\notin S$. Similarly, $sg\notin S$.
$\forall s\in S,sg\notin S \Rightarrow Sg\cap S = \emptyset$, and this lead to a contradiction with proposition 2!
Q.E.D.
A: As I wrote in the comments I don't understand how you get the main result from Proposition 2. However, here is a different way of getting the main result from proposition 2 (proposition 2 of the answer, not the question).
I use the notation of your answer not your question, so the 'special' element is called $g$, not $a$.
Let $s_1 \in S \cap Sg$, it exists by Proposition 2.
Let $s_2 \in S$ be such that $s_1 = s_2g$.
Let $B = \{e, g\}$. The premise that $S \cup gS = G$ is equivalent to $BS = G$.
Let $B' = s_2B$. Then on one hand we have:
$B'S = s_2BS = s_2G = G$
On the other hand we have $B' = \{s_2, s_2g\} = \{s_2, s_1\} \subset S$
So $B' \subset S$ and hence $G = B'S \subset SS$ as we wanted to show.
I already had thought up this argument a few days ago, but had no use for it because I couldn't prove Proposition 2. So I was really enthusiastic to read your elegant proof of Proposition 2 in the answer. This question was becoming a bit of an obsession, so I am happy it is settled now.
