# Thread: Proof involving Cosets

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 $h\in H$" in addition to "for all $a\in G$".

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

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

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