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I cannot see how to find a way to prove that if $H$ is a subgroup of $G$ such that the product of two right cosets of $H$ is also a right coset of $H,$ then $H$ is normal in $G.$

(This is from Herstein by the way.)

Thank you.

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    How to prove HaHb = Hc then HaHb = Hab2018-09-19

3 Answers 3

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Hint: if $HaHa^{-1}$ and $Ha^{-1}Ha$ are right cosets they must be $H$ because they contain the identity.

(I have updated my hint to involve both $HaHa^{-1}$ and $Ha^{-1}Ha$ because $aHa^{-1}\subseteq H$ is not by itself equivalent to $aHa^{-1}=H$ when $H$ is infinite; see counterexamples here, here, here.)

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    @Hansie $HaHb$ contains $eaeb$, which is $ab$, and thus $HaHb$ is $Hab$. (If $g$ is any element of a right coset $X$ then $X=Hg$.)2018-09-20
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A (not quite as) short alternate proof:

If $HaHb=Hc$ then $HaHb=Hab$. @anon's short proof chooses $b=a^{-1}$, but you can also choose $b=1$, since $HaH = Ha \iff 1aH \subseteq Ha$

Of course to get equality, we also have to use $Ha^{-1}H =Ha^{-1} \iff a^{-1} H \subseteq Ha^{-1} \iff Ha \subseteq aH $


In general, $HaHb=Hab \iff aHb \subseteq Hab$, so if we want $aH=Ha$ we choose $b=1$ and if we want $aHa^{-1}= H$ we choose $b=a^{-1}$. If groups are finite, we don't even have to pay attention to $\subseteq$ versus $=$.

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    How to prove HaHb = Hc then HaHb = Hab2018-09-19