I see there is an old string addressing this question, but I need to prove it using set containment both directions. Not sure where to start.

a, b are in G, a group. H is a subgroup. If aH = bH then Ha^(-1) = Hb^(-1).

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- Apr 16th 2008, 05:30 AMGoldendoodleMomAbstract Algebra Problem
I see there is an old string addressing this question, but I need to prove it using set containment both directions. Not sure where to start.

a, b are in G, a group. H is a subgroup. If aH = bH then Ha^(-1) = Hb^(-1). - Apr 16th 2008, 06:51 AMIsomorphism
You can do this easily if you write the conditions clearly and think reverse.

To prove: $\displaystyle Ha^{-1} \subset Hb^{-1}$

That is to prove: $\displaystyle \forall h \in H, ha^{-1}b \in H$

But that is the same as proving $\displaystyle a^{-1}b \in H$

Now since, $\displaystyle aH = bH \Rightarrow a^{-1}b \in H$, we are done.

The other way round is exactly similar except for different variable name.