If $aH = bH$ then $Ha^{-1} = Hb^{-1}$, prove or find a counter example.

$H$ is a subgroup of $G$ and $a,b$ are in $G$.

I was rather lost for a while but I think I may have actually proved it, though I am unsure… Does this make sense?

\begin{aligned} aH &= bH \\ &\implies b \in aH \\ &\implies \exists h \in H : ah = b \\ &\implies a^{-1}ah = a^{-1}b \\ &\implies h = a^{-1}b \\ &\implies hb^{-1} = a^{-1}bb^{-1} \\ &\implies hb^{-1} = a^{-1} \\ \end{aligned}

This means $a^{-1}$ is an element of $Hb^{-1}$, so they are equal.

I wasn't quite sure I could do those operations. Can anyone tell me where to learn to do the special symbols, like quantifiers and relations because it would make it simpler.

$aH=bH$ does not imply that $a$ or $b \in H$. It just implies they are in the same coset. It does imply that $b^{-1}a \in H$

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As to your argument, it is (mostly) correct. My only quibbles would be of presentation. From $aH=bH$ you conclude that there must exist $h\in H$ such that $b=ah$. That's fine. Your next step should be that "there exists $h\in H$ such that $a^{-1}b=a^{-1}ah = h$". Then "There exists $h\in H$ such that $a^{-1}=a^{-1}bb^{-1}=hb^{-1}$", hence $a^{-1}$ is in the coset $Hb^{-1}$. Your final statement "hence they are equal" is technically wrong, since $a^{-1}$ is not equal to the coset $Hb^{-1}$; what you meant to say is that the since $a^{-1}$ is in $Hb^{-1}$, then the coset $Ha^{-1}$ is equal to the coset $Hb^{-1}$. Other that these quibbles, the argument seems correct to me.

Now, to see if you understand it well, try to figure out which, if any, of your implications are reversible, to see if the converse also holds.

You're on the right track. $aH = bH$ is equivalent to $a^{-1}b \in H$, and $Ha^{-1} = Hb^{-1}$ is equivalent to $a^{-1} (b^{-1})^{-1} = a^{-1}b \in H$, which is the same thing. One approach that should work would be to turn your arguments into statements of this nature.