Suppose $H_1$ and $H_2$ are subgroups of $G$. Suppose $a_1$ and $a_2$ are two elements of $G$ such that $a_1H_1 = a_2H_2$. Prove that $H_1 = H_2$.

Question

I believe that the problem stated in the title is false. I have attempted to prove it by manipulating the assumption that there is some $h_1 \in H_1$ and some $h_2 \in H_2$ such that $a_1h_1 = a_2h_2$. The idea was to come to the conclusion that $H_1 \subset H_2$ and $H_2 \subset H_1$ showing they were equal. After trying every type of manipulation I could think of I could not come to this conclusion.

Despite this, I cannot seem to find a counter example. I've put some different GPTs on ChatGPT through the wringer trying to find one or some proof but it also has fell short. Anyone have any ideas?

Answer 1

The statement is correct. We have $a_2^{-1}a_1H_1=H_2$. This implies that the coset $a_2^{-1}a_1H_1$ contains the identity of $G$, and so it has to be equal to the coset $H_1$. Hence $H_1=a_2^{-1}a_1H_1=H_2$.