If $|A| > \frac{|G|}{2} $ then $AA = G $ [closed]

I'v found this proposition.

If $G$ is a finite group , $ A \subset G $ a subset and $|A| > \frac{|G|}{2} $ then $AA = G $.

Why this is true ?

Solution 1:

it is true.

let $A^{-1}=\{a^{-1}|a\in A\}$ then notice that $|A^{-1}|=|A|$ and let $x\in G$.

Now we can say that $|xA^{-1}|=|A|$ .Since $2|A|>|G|$ then $xA^{-1}$ and $A$ must intersect,otherwise we have contradiction.

Thus,we must have $xa_1^{-1}=a_2$ for some $a_1^{-1}\in A^{-1}$ and $a_2\in A\implies$ $x=a_1a_2\in AA$ we are done.