1. ## Proof involving Cosets

Hey everyone I have a homework problem me and a few friends have been working on for over a week and cant find a solution we all like. Any input would be greatly appreciated. Heres the Problem:

Let H be a subgroup of G with aha(inverse)∈H for all a∈G. Prove that
aH = Ha for all a∈G.

2. Let H be a subgroup of G with aha(inverse)∈H for all a∈G.
You probably mean "for all $\displaystyle h\in H$" in addition to "for all $\displaystyle a\in G$".

This is easy. To show $\displaystyle aH\subseteq Ha$ for all $\displaystyle a\in G$ you need to show that for every $\displaystyle a\in G$ and every $\displaystyle h\in H$ there exists $\displaystyle h'\in H$ such that $\displaystyle ah=h'a$. An equivalent equality can be obtained by multiplying both sides on the right by $\displaystyle a^{-1}$. Does such $\displaystyle h'\in H$ actually exist?

For the opposite inclusion, again write the statement to be proved and multiply the equality by $\displaystyle a^{-1}$ on the left. Also, $\displaystyle a^{-1}$ can be renamed into some $\displaystyle a_1\in G$.

3. thank you so much i seem to always make these 10x harder than they really are. I really appreciate it