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.

Solution 1:

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.)

Solution 2:

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 $=$.

Solution 3:

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.