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).
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).
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.