I'm studying a topic in group theory, I'm stuck with the following point.

$G$ is a finite group, $A \subset G $ a subset. We have $ H= AA^{-1} = A^{-1}A $, and we know that $$|H| \leq \frac{1}{1- \varepsilon}|A| \ ,\ \ \ \ \varepsilon > 0.$$

Then if $\varepsilon < \frac{1}{2} $, given $x , y \in H $ there are representations $x = dc^{-1}$ and $y = ef^{-1}$ with $c = e$.

Why this is true ?


Notice that $|H|\lt 2|A|$, hence $$|xA\cup yA|\gt 2|H|-|xA\cap yA|.$$ Since $|H|\geqslant |A|$ and $|xA\cup yA|\leqslant 2|A|$ we have $xA\cap yA\neq \emptyset$.

  • $\begingroup$ sorry but i don't understand why your inequality implies $xA \cap yA \neq \varnothing $ $\endgroup$ – WLOG Feb 23 '14 at 22:08
  • $\begingroup$ See edit (there was a typo). $\endgroup$ – Davide Giraudo Feb 23 '14 at 22:11

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