# Is it true that $aH = bH$ iff $ab^{-1} \in H$

Let $H$ be a subgroup of a group $G$. The I know that for $a,b\in G$ we have $aH = bH$ if and only if $a^{-1}b \in H$.

My question is if it is also true that $aH = bH$ if and only if $ab^{-1} \in H$? Does it matter where the $-1$ goes?

I am guessing that this is not true since we keep putting the $-1$ on the left element, but I am also thinking that it might be true because they look like the same.

If $G$ is abelian, then it doesn’t matter, so a possible counter example would have to involve a non-abelian group. If $H$ is a normal subgroup, then it should also be true because then $a^{-1}bH = Ha^{-1}b$.

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No: here’s a counterexample. Let $G=S_3, H=\langle (1,2)\rangle$,$a=(1,2,3)$, $b = (1,3)$. Then $a^{-1}b = (1,2) \in H$ so $aH= bH$, but $ab^{-1} = (1,2,3)(1,3) = (2,3) \notin H$.

(I multiply permutations from right to left here).

No. This is why, in detail:
\begin{align*}aH=bH&\iff b^{-1}(aH)=b^{-1}(bH)\iff (b^{-1}a)H=(b^{-1}b)H=H\\&\iff b^{-1}a\in H \end{align*}

Now if it were true that, for all $a,b\in G$,
$$aH=bH\iff ab^{-1}\in H$$
it would imply that $aH=bH\iff Ha=Hb$ for all $a,b$. In particular, setting $b=e$, it would imply that for all $a\in H$, $aH=Ha$, i.e. $H$ would be a normal subgroup.