# Thread: showing identity in proving k ={x in g|x = aha^-1 h in H}

1. ## showing identity in proving k ={x in g|x = aha^-1 h in H}

Ok the problem is let H subgroup of G, a fixed in G, let K = {x in G|x = aha^-1 h in H} Prove or prove K subgroup of G. Im pretty sure i got closure right and the identity is pretty easy but im stuck on showing inverse in K. We know for some x in G x = aha^-1 and we know x^-1 in G because G group. but would that transfer to x^-1 = (aha^-1)^-1 or what that still doesnt seem to get me where i need to go

**Sorry about the title i just noticed i put identity when i meant inverse

2. Originally Posted by ChrisBickle
Ok the problem is let H subgroup of G, a fixed in G, let K = {x in G|x = aha^-1 h in H} Prove or prove K subgroup of G. Im pretty sure i got closure right and the identity is pretty easy but im stuck on showing inverse in K. We know for some x in G x = aha^-1 and we know x^-1 in G because G group. but would that transfer to x^-1 = (aha^-1)^-1 or what that still doesnt seem to get me where i need to go

**Sorry about the title i just noticed i put identity when i meant inverse

For any $m\in\mathbb{Z}\,,\,\,(aha^{-1})^m=ah^ma^{-1}$. You also may want to use $(xy)^{-1}=y^{-1}x^{-1}$ , which is true in any group.

Tonio