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

1. ## 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

2. 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 $aha^{-1}bhb^{-1}$ necessarily of this form? Of course not!

So, think any non-abelian group and try to find a counter-example. Start with $S_3$.

3. Originally Posted by Swlabr
Is $aha^{-1}bhb^{-1}$ necessarily of this form? Of course not!

So, think any non-abelian group and try to find a counter-example. Start with $S_3$.
You misread the problem. a is fixed

4. 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 $(aha^{-1})(aga^{-1}) \in K$?

What do you (rainyice) think $(hah^{-1})^{-1}$ is?

5. Originally Posted by Swlabr
Ooooh, that's boring!

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

Is $(aha^{-1})(aga^{-1}) \in K$?

What do you (rainyice) think $(hah^{-1})^{-1}$ is?

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

6. Originally Posted by rainyice
I think it will become ha^(-1)h^(-1)
Precisely, and so you have your result!

7. Originally Posted by Swlabr
Precisely, and so you have your result!

so it isn't a subgroup?

8. It is. Swlabr had a typo there. He meant $aha^{-1}$, and your result should be $ah^{-1}a^{-1}$, 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.