Product of right cosets equals right coset implies normality of subgroup 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.
 A: 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 $=$.
A: Suppose g ∈ G and consider the product of two right cosets HgHg−1
. Observe
egeg−1 = e ∈ HgHg−1 and since by hypothesis HgHg−1
is a right coset, it would have
to be the right coset H = He as right cosets are either equal or disjoint and in this case
e ∈ H = He so we must have equality HgHg−1 = H. Therefore, for every g ∈ G we have
HgHg−1 = H so clearly gHg−1⊆ HgHg−1 = H. Hence, gHg−1 ⊆ H for all g ∈ G so H
is a normal subgroup of G.
A: 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.)
