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Let $G$ be a group and $H$ a subgroup. Consider $H/G\backslash H$ the set of double cosets of $H$ in $G$. Show $H$ is normal in $G$ if and only if the mapping $f$ from $G/H$ into $H/G\backslash H$ defined by $f(aH) = HaH$ for all $a$ in $G$ is a bijection. The part of $H$ normal in $G$ implies a mapping $f$ is a bijection is easy. (It was a comment in Bourbaki Algebra).

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If $H$ is not normal, then there exist $a\in G$, $h \in H$ with $a^{-1}ha \not\in H$, so $ha \not\in aH$ and hence $aH \ne haH$. But $f(aH) = f(haH)$, so $f$ is not a bijection.

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