If $aH=bH \implies Ha=Hb$ for a subgroup H having *finite index*, then $gH=Hg$ for all $g \in G$?

Question

Problem 2.5.9 of Herstein's Topics in Algebra asks us to prove that if $H$ is a subgroup of a group $G$ such that $Ha \not = Hb \implies aH \not = bH$, then $gHg^{-1} \subset H$ for all $g \in G$, which is equivalent to $gH \subset Hg.$ Suppose I have proved this (this result is of course true). I want to prove more, namely that, under the same premise, $gH = Hg$ for all $g \in G$. I think I can prove it if I further assume that the index of $H$ in $G$ is finite. Is this true? And if so, is this additional hypothesis necessary as well as sufficient? I.e. can you give an example of a subgroup $H$ of infinite index such that $aH \subset Ha$ but not vice versa? Thanks in advance. [DISCLAIMER: I am not yet familiar with normal subgroups]

Here's my attempt at a proof:

Suppose that $|G:H|=n \in \mathbb{N}$ and that $aH \subset Ha$ for all $a \in G$. Now, suppose that there is an element $x$ such that $x \in Ha$ but $x \not \in aH$. Then $x$ must be contained in a different left coset, say $bH$, because the left cosets form a partition of $G$. Then, by our hypothesis, $x \in Hb$, which implies $Hb = Ha$, because the right cosets form a partition of $G$ as well. So far we have shown that $aH, bH \subset Ha=Hb$. But now we can prove that there are more left cosets than right ones! In fact, the remaining $n-1$ right cosets distinct from $Ha$ must contain at least one left coset, by our hypothesis, but $Ha$ contains two left cosets. So there are at least $n+1$ left cosets, a contradiction. Therefore $aH=Ha. \square$

Answer 1

Your proof seems correct to me.

In fact, you can remove the finite index hypothesis. If $gHg^{-1} \subset H$ for all $g\in G$, then for each $g\in G$ you also have $$ H = g^{-1}(gHg^{-1})g \subset g^{-1}Hg = g^{-1}H(g^{-1})^{-1} \subset H, $$ so we must have equality throughout, i. e. $H = g^{-1}Hg$ (for each $g\in G$).

More generally, for any subgroup $H$ of $G$ the following holds: the set $N:= \{ g\in G \mid gHg^{-1}\subset H\}$ is a subgroup of $G$ if and only if $gHg^{-1} = H$ for all $g\in N$. (In that case $N$ is the normalizer of $H$ in $G$.)

Answer 2

You don't need any additional hypotheses. Multiplying both sides by $g^{-1}$ on the left and $g$ on the right yields

$$gHg^{-1} \subset H \implies H \subset g^{-1}Hg$$

Since this is true for all $g \in G$, we can substitute $g$ for $g^{-1}$, concluding

$$gHg^{-1} \subset H \subset gHg^{-1}$$

$$H = gHg^{-1}$$

This is equivalent to

$$gH = Hg$$