A property for infinite groups

Question

Let $G$ be an infinite group, $F$ a finite subset of $G$ and $A=G\setminus F$. Is it true that $A^{-1}A=AA^{-1}=G$ (what about $AA=G$)?

($A^{-1}=\{ a^{-1}:a\in A\}$)

Answer 1

We have $A^{-1}A=AA^{-1}=AA=G$.

Show just $A^{-1}A=G$, the others are similar.

Suppose there is a $g\in G$ with $g\notin A^{-1}A$. Then we have for every $a\in A$ that $g\notin a^{-1}A$ and so $ag\notin A$. So we have $Ag\cap A=\emptyset$ and so $Ag\subseteq G\setminus A$. This is contradiction to $|G\setminus A|<\infty$ since $A\to Ag, g\mapsto ag$ is bijective and so $|A|=|Ag|=\infty$.