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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 ?

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closed as off-topic by Did, Ittay Weiss, Adam Hughes, user98602, Kevin Carlson Jan 15 '15 at 23:23

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  • 1
    $\begingroup$ Found it where? Presumably, we're talking finite groups. $\endgroup$ – Thomas Andrews Feb 23 '14 at 22:37
  • $\begingroup$ We had this before on math.stackexchange, but I don't know how to fed the search function. $\endgroup$ – Martin Brandenburg Feb 23 '14 at 22:44
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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.

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  • $\begingroup$ In your very last implication, you're using the fact that $A$ is closed under miltiplication. Nevertheless $A$ is only a subset, so... $\endgroup$ – Gabriel Romon Mar 6 '14 at 11:38
  • $\begingroup$ $A$ is not closed under multiplication. $\endgroup$ – mesel Mar 6 '14 at 12:10
  • $\begingroup$ Oh it's $AA$! I'm sorry, $A$ prints as a blank character on my phone so the whole time I was considering $AA$ as $A$... $\endgroup$ – Gabriel Romon Mar 6 '14 at 12:24
  • $\begingroup$ @GabrielR. :it is okey. $\endgroup$ – mesel Mar 6 '14 at 12:43

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