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

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 $$aH=Ha. \square$$

• Compare with this duplicate and its answers. – Dietrich Burde Mar 14 at 12:00
• @DietrichBurde The answers in your linked question only prove $gHg^{-1} \subset H$ and not equality. So I don't think they address the OP’s problem. – Claudius Mar 14 at 12:15
• @Claudius This is not true. They prove equality, e.g., see Deven's answer. – Dietrich Burde Mar 14 at 12:18
• @DietrichBurde Yes, Deven claims to have shown $gHg^{-1} =H$. But what he really did was to prove $gHg^{-1} \subset H$. At least, I don't see how he proved the other inclusion. – Claudius Mar 14 at 12:20

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 $$G$$.)

• That's brilliant, thank you. Now I feel stupid for not having come up with that immediately! But why would Herstein keep this result from the reader and only ask to prove one inclusion? And what about this thread math.stackexchange.com/questions/217601/… in which there seems to be a counterexample? – The Footprint Mar 14 at 12:36
• In the answer of Brian M. Scott the set $\{g\in G\mid gHg^{-1} \subset H\}$ is not stable under inversion, hence not a group. It is only closed under composition. – Claudius Mar 14 at 12:40

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

$$gH = Hg$$