Thread: Prove or disprove that K is a subgroup of G

Let H be a subgroup of G, let a be a fixed element of G, and let K be the set of all elements of the form a*h*a^(-1), where h E H. That is, K = {x E G | x= a*h*a^(-1) for some h E H}
Prove or disprove that K is a subgroup of G

Originally Posted by rainyice

Let H be a subgroup of G, let a be a fixed element of G, and let K be the set of all elements of the form a*h*a^(-1), where h E H. That is, K = {x E G | x= a*h*a^(-1) for some h E H}
Prove or disprove that K is a subgroup of G

Is necessarily of this form? Of course not!

So, think any non-abelian group and try to find a counter-example. Start with .

Originally Posted by Swlabr

Is necessarily of this form? Of course not!

So, think any non-abelian group and try to find a counter-example. Start with .

You misread the problem. a is fixed

Originally Posted by FancyMouse

You misread the problem. a is fixed

Ooooh, that's boring!

You then just verify that such a set is, in fact, a subgroup.

Is ?

What do you (rainyice) think is?

Originally Posted by Swlabr

Ooooh, that's boring!

You then just verify that such a set is, in fact, a subgroup.

Is ?

What do you (rainyice) think is?

I think it will become ha^(-1)h^(-1)

Originally Posted by rainyice

I think it will become ha^(-1)h^(-1)

Precisely, and so you have your result!

Originally Posted by Swlabr

Precisely, and so you have your result!

so it isn't a subgroup?

It is. Swlabr had a typo there. He meant , and your result should be , which shows that every element in K has an inverse.
In fact K is just a conjugate subgroup of H. You'll know more about it once you learned group action on sets and Sylow theorems.