TITLE: Evaluating the proof for the following statement: If $aH=Ha$, then $a^{-1}H=Ha^{-1}$
QUESTION [6 upvotes]: From Pinter's "A Book of Abstract Algebra", Chapter 13 Exercise E4 asks to prove the following, given that $H$ is a subgroup:

If $aH=Ha$, then $a^{-1}H=Ha^{-1}$

At first I thought that this was a super straightforward proof and solved it as follows:
let $x \in H$
Then 
$ax=xa \implies a^{-1}ax=a^{-1}xa \implies x=a^{-1}xa \implies xa^{-1}=a^{-1}xaa^{-1} \implies xa^{-1}=a^{-1}x$ 
However, I then reconsidered what the antecedent of the "implication to prove" was really saying. From what I understand, "if $Ha=aH$..." is really saying these sets are equal...i.e. these sets contain the same elements. This antecedent is not necessarily stating that $a$ commutes with all elements of $H$. Because of this, should I rewrite my proof more generally in the form of:
$x,y \in H$ and then proceed with $ax=ya$...before getting it into the final form of $xa^{-1}=a^{-1}y$. Further, if this is correct, can I now conclude that $a^{-1}H=Ha^{-1}$?
Cheers~

REPLY [4 votes]: Your second interpretation is the correct one.
I guess, $H$ is assumed to be a subgroup, so that in particular 
$$H^{-1}:=\{h^{-1}\mid h\in H\} = H$$
and thus, inverting every element of both sets, from $aH=Ha$, we conclude $H^{-1}a^{-1}=a^{-1}H^{-1}$, so these combine to $Ha^{-1}=a^{-1}H$.