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The question asks

(a) Let $A$ be a subset of finite group $G$ with strictly greater than $|G|/2$ elements. Show $AA=G$
and
(b) Show this can fail in a monoid.

I've been working on this for awhile but am not sure what to do...

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closed as off-topic by Adam Hughes, Daniel W. Farlow, Kushal Bhuyan, Claude Leibovici, user99914 Jul 19 '16 at 4:43

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  • $\begingroup$ (a) is solved here. $\endgroup$ – user26857 Jul 19 '16 at 8:31
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(a) Let $g\in G$ and $A^{-1}:=\{a^{-1}|a\in A\}$. Then we know that $|gA^{-1}|=|A^{-1}|=|A|$, so $|gA^{-1}|+|A|>|G|$. Thus, $gA^{-1}$ and $A$ have a common element, and...

(b) Consider $(\mathbb{Z}/6\mathbb{Z}, \times)$. Let $A=\{0, 2, 3, 4\}$.

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