Let $A,B$ be two subsets of a finite group $G$. If $|A|+|B|>|G|$, show that $G=AB$

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.

• When you say $A$ and $B$ are subsets, do you actually mean subgroups? Feb 14, 2013 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 $AB$ Feb 14, 2013 at 15:12
• That $AB\subseteq G$ is trivial. Feb 14, 2013 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. Feb 14, 2013 at 16:06
• @peoplepower, why should this be a counterexample? Surely $\mathbb Z_5 = A + B$ here. Feb 14, 2013 at 16:34

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 $ab = a a^{-1} g = g$.
• +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. Feb 14, 2013 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 $b=a^{-1} g$?" Feb 14, 2013 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. Apr 10, 2014 at 20:42