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 $AA=G$)?

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


closed as off-topic by Derek Holt, John B, Shailesh, Silvia Ghinassi, Daniel W. Farlow Mar 2 '16 at 2:41

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    $\begingroup$ Yes (for all questions). This is essentially the same argumnet as in math.stackexchange.com/questions/687715 $\endgroup$ – Derek Holt Mar 1 '16 at 18:03
  • $\begingroup$ Why?, note that the hypothesis of this question are completely different !(but with the same sentence) $\endgroup$ – M.H.Hooshmand Mar 1 '16 at 18:12
  • $\begingroup$ @M.H.Hooshmand True, but the argument is exactly the same: the cardinality of $A$ is strictly larger than that of $F$, so there must be an element in $A$ that is not disqualified by any element of $F$. We're applying the same argument, not the same theorem. $\endgroup$ – Erick Wong Mar 1 '16 at 19:20

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 $|A|=|Ag|=\infty$.

  • $\begingroup$ This works even if G is finite, provided that |G|>2|F|. $\endgroup$ – DanielWainfleet Mar 1 '16 at 19:37

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