Subgroup conjugation and cosets

[ttm7nana]

Hello, I am having trouble with the following problem.

Suppose that H is a subgroup of G such that whenever Ha≠Hb then aH≠bH. Prove that gHg^(-1) is a subset of H.

I have tried to manipulate the following equation for some ideas

H = Hgg^(-1) = gg^(-1)H

but I don't know how to go from here. I can't figure out how I can use the conditions to show that every element in gHg^(-1) is also in H.
I also know that gHg^(-1) is a subgroup of G, but I don't know if this fact can be used here.
It well be great if someone can point me in the right direction. Thank you.

 

[Dick]

Try it this way. Can you use "Ha≠Hb then aH≠bH" to show that [itex]ab \not \in H[/itex] iff [itex]ba \not \in H[/itex]? Apply that to ghg^(-1) where h is an element of H. Oh, and welcome to the forums!

 

[ttm7nana]

Hi, thanks for your help and the welcome!
I think I've got it. Can you check my work?

Let ab[itex]\notin[/itex]H. From this we get Ha ≠Hb^(-1). Using Ha≠Hb then aH≠bH, aH ≠ b^(-1)H. This implies that ba[itex]\notin[/itex]H. And the other direction is the same procedure. Therefore ab[itex]\notin[/itex]H iff ba[itex]\notin[/itex]H

Now assume for a sake of contradiction that gHg^(-1)is not a subset of H. Then we get ghg^(-1)[itex]\notin[/itex]H for some h[itex]\in[/itex]H. Then from ab[itex]\notin[/itex]H iff ba[itex]\notin[/itex]H, we get g^(-1)gh=h[itex]\notin[/itex]H. Which is a contradiction because h[itex]\in[/itex]H. Therefore gHg^(-1)[itex]\subset[/itex]H.

I'm not very confident about the first half of the proof.

 

[Dick]

Hi, thanks for your help and the welcome!
I think I've got it. Can you check my work?

Let ab[itex]\notin[/itex]H. From this we get Ha ≠Hb^(-1). Using Ha≠Hb then aH≠bH, aH ≠ b^(-1)H. This implies that ba[itex]\notin[/itex]H. And the other direction is the same procedure. Therefore ab[itex]\notin[/itex]H iff ba[itex]\notin[/itex]H

Now assume for a sake of contradiction that gHg^(-1)is not a subset of H. Then we get ghg^(-1)[itex]\notin[/itex]H for some h[itex]\in[/itex]H. Then from ab[itex]\notin[/itex]H iff ba[itex]\notin[/itex]H, we get g^(-1)gh=h[itex]\notin[/itex]H. Which is a contradiction because h[itex]\in[/itex]H. Therefore gHg^(-1)[itex]\subset[/itex]H.

I'm not very confident about the first half of the proof.

I think you've got it. The justification for the first part is that if [itex]c \notin H[/itex] iff [itex]cH \neq H[/itex]. You might want to explicitly mention that.

 

[ttm7nana]

Okay, thank you very much!

If I may ask, what was your thought process in first showing ab[itex]\notin[/itex]H iff ba[itex]\notin[/itex]H? I would have never thought of that step on my own.

 

[Dick]

Okay, thank you very much!

If I may ask, what was your thought process in first showing ab[itex]\notin[/itex]H iff ba[itex]\notin[/itex]H? I would have never thought of that step on my own.

Sure. If you skip to the end the question is whether ghg^(-1) is 'in H or not'. So then I looked at Ha≠Hb and tried to figure out what that was telling me about something being 'in H or not'. And, of course, it's telling you ab^(-1) 'is not in H'. aH≠bH says b^(-1)a is not in H. So that looks like it's telling me if the product of two elements is not in H, then I can commute them and it's still not in H. Etc.

 

[ttm7nana]

Okay, that makes sense.
Thank you again for your help!