# Evaluating the proof for the following statement: If $aH=Ha$, then $a^{-1}H=Ha^{-1}$

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~

• oh, thank you! I'll make the edit now. – S.Cramer Dec 6 '19 at 1:07

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 $$Ha^{-1}=a^{-1}H$$.
• Just because I have never seen that notation before...I wanted to confirm that I am understanding it correctly. The inverse of the set H (i.e. $H^{-1}$) is effectively equal to the set H. This is because H is a group and therefore all inverse elements are already contained. Is that correct? – S.Cramer Dec 6 '19 at 1:15