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

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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 $A^{-1}A=AA^{-1}=G$?

## marked as duplicate by Derek Holt group-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 9 '16 at 7:38

• What operation are you representing by concatenation here? Does $AA^{-1} = \{ ab : a \in A, b \in A^{-1} \}$ – Q the Platypus Feb 9 '16 at 6:40
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 $g=ca^{-1}\in AA^{-1}.$