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This question already has an answer here:

Let $G$ be a finite group, $A\subseteq G$ and put $A^{-1}=\{ a^{-1}:a\in A\}$.

Is it true that if $|A|>\frac{|G|}{2}$ then $A^{-1}A=AA^{-1}=G$?

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marked as duplicate by Derek Holt group-theory Feb 9 '16 at 7:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What operation are you representing by concatenation here? Does $ AA^{-1} = \{ ab : a \in A, b \in A^{-1} \} $ $\endgroup$ – Q the Platypus Feb 9 '16 at 6:40
  • $\begingroup$ Obviously, yes. $\endgroup$ – M.H.Hooshmand Feb 9 '16 at 6:42
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    $\begingroup$ See this question, the proof is nearly identical. $\endgroup$ – Jendrik Stelzner Feb 9 '16 at 6:55
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Take any $g\in G$ then $|gA|=|A|>0.5|G|$ thus $(gA)\cap A\neq \emptyset .$ Let $c\in (gA)\cap A$ then $c=ga$ and therefore $g=ca^{-1}\in AA^{-1}.$

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