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Let $A,B$ be two subsets of a finite group $G$. If $|A|+|B|>|G|$, show that $G=AB$. My attempt is : Since $|A|+|B|>|G|$, there exists one common element in both sets $A$ and $B$, say $g$. Then since $G$ is a group, by closure, $g^2 \in G$, which implies that $G \subset AB$. Let $a \in A$, $b \in B$. Then I get stuck at proving another inclusion.

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When you say $A$ and $B$ are subsets, do you actually mean subgroups? – anon271828 Feb 14 '13 at 15:12
Inclusion $AB\subset G$ is obvious. But your proof is wrong since you proved only that some element of $G$ is in $AB$ – Norbert Feb 14 '13 at 15:12
That $AB\subseteq G$ is trivial. – Tobias Kildetoft Feb 14 '13 at 15:13
@anon271828 If $A$ and $B$ were subgroups there'd be nothing to do. Since $|A|+|B|>|G|$ implies either $|A|>|G|/2$ or $|B|>|G|/2$ which would force either $A=G$ or $B=G$. Although it's possible the question is that straightforward. – JSchlather Feb 14 '13 at 16:06
@peoplepower, why should this be a counterexample? Surely $\mathbb Z_5 = A + B$ here. – Andreas Caranti Feb 14 '13 at 16:34
up vote 9 down vote accepted

Take any $g \in G$. Let $A^{-1} = \lbrace a^{-1}, a \in A \rbrace$. Then $\vert A^{-1}g \cap B \vert \gt 0$ by easy counting. Let $b = a^{-1} g$ for some $a \in A$. Then $ab = a a^{-1} g = g$.

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+1 It's the argument that can be used to show that in a finite field $F$ of odd order $q$ every element is the sum of two squares. In this case $A = B = \{ u^2 : u \in F \}$ has $(q+1)/2 > q/2$ elements. – Andreas Caranti Feb 14 '13 at 16:30
"Let $b=a^{-1} g, \exists a \in A$" Ok it's clear what you mean, but this really contradicts all rules of mathematical well-defined formulas. Why not "Choose $a \in A$ with $b=a^{-1} g$?" – Martin Brandenburg Feb 14 '13 at 16:35
Good proof. Two more corrections: You can say $A^{-1}g\cap B\ne\varnothing$, $\left\vert A^{-1}g \cap B \right\vert \ne 0$, or even $\left\vert A^{-1}g \cap B \right\vert \gt 0$ but not $\left\vert A^{-1}g \cap B \right\vert \ne \varnothing$. And using \phi$(\phi)$ instead of \emptyset$(\emptyset)$ or \varnothing$(\varnothing)$ for the empty set is just bad juju. – Travis Bemrose Apr 10 '14 at 20:42

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