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 $x \in H$


$ax=xa \implies a^{-1}ax=a^{-1}xa \implies x=a^{-1}xa \implies xa^{-1}=a^{-1}xaa^{-1} \implies xa^{-1}=a^{-1}x$

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 $a^{-1}H=Ha^{-1}$?


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

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 $Ha^{-1}=a^{-1}H$.

  • $\begingroup$ 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? $\endgroup$ – S.Cramer Dec 6 '19 at 1:15
  • 1
    $\begingroup$ @S.Cramer yes, that's correct. $\endgroup$ – Berci Dec 6 '19 at 1:25

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