Subgroups - Conjugates

1. Jul 29, 2014

Justabeginner

1. The problem statement, all variables and given/known data

Prove that the set A= {a in G|aHa^-1=H} is a subgroup of G.

2. Relevant equations
H and K are conjugates of a group G with a in G, where: aHa^-1= {aha^-1| h in H}= K.

3. The attempt at a solution
For some x,y in aHa^-1, h1,h2 in H also exists where:
X= ah1a^-1 and Y=ah2a^-1

Then, XY= ah1h2a^-1 which is contained in aHa^-1.

And x^-1= ah1^-1a^-1 which is in aHa^-1.

aHa^-1 is a subgroup of G. Since A contains a in G such that aHa^-1=H, and aHa^-1 is a subgroup of G, then A is also a subgroup of G.

Is this logic correct? Thanks!

Last edited: Jul 29, 2014
2. Jul 29, 2014

I can't really help, but just out of curiosity, what is $H$ here?

3. Jul 29, 2014

Justabeginner

Oops, I failed to mention H and K are conjugates of a group G with a in G, where: aHa^-1= {aha^-1| h in H}= K.

Thank you for bringing this to my attention!

4. Jul 29, 2014

gopher_p

Everything looks correct until the very last sentence. At the very least, you need to flesh that claim out a bit more.

If I may, I'd like to suggest you take a completely different approach to the problem and show, given $a,b\in A$, that $ab\in A$ and $a^{-1}\in A$. I.e. instead of showing that $aHa^{-1}$ is a subgroup and then using that to prove the original claim, just cut straight to the claim and show directly that $A$ is a subgroup.

Last edited: Jul 29, 2014
5. Jul 29, 2014

Justabeginner

Thank you for the different approach.

I will try it out, so please let me know if the following logic and assumptions are correct.

If a,b in A, then by closure and inverse property of a group, ab in A, and a^-1 in A.
So then (aH)a^-1 in A if a^-1 in A.
Since aHa^-1= H, then H is a subgroup of A, and because H is a subgroup of G, A is a subgroup of G?
That last part does not seem right to me, though.

6. Jul 29, 2014

gopher_p

You're meant to demonstrate that $A$ is a subgroup. It appears as though you're assuming $A$ is a subgroup from the outset. You're supposed to explain why $A$ is closed under the group operation and inverses of $G$. The only thing you're allowed to assume is the definition of $A$ and that $H$ is a subgroup of $G$.

$H$ is a subgroup of $A$, and that follows from the facts that $H$ is a group, which is an assumption of the problem, and that $H$is a subset of $A$, which you'd need to demonstrate.

Alright, so given $a,b\in A$, you want to show that $ab\in A$. In order for $ab$ to be a member of $A$ it must be true that $abH(ab)^{-1}=H$, right? Can you show that? Remember that you're assuming that $a$ and $b$ are members of $A$, and you want to use that fact and the fact that $H$ is a subgroup of $G$, and those facts only, to show that $ab\in A$.

Also, it seems like you're starting to resort to stabs in the dark here (that you have a question mark at the end of your proof is telling). I would highly recommend that you do not make a claim that you cannot justify. For instance, in your first sentence, you claim that $A$ has the "closure and inverse property of a group". Why? If you can't answer "Why?", then you need to either fill in some gaps or wad up the scratch paper and start over.

7. Jul 29, 2014

Justabeginner

If a, b in A and H is a subgroup of G (assumption of entire problem) then abH(ab)^-1 in G too. So, abHb^-1a^-1 in G. This means that a(bHb^-1)a^-1 in G, and if bHb^-1 in G, then b in G. Also, aa^-1 in G is equivalent to e in G so A in G.

I will try to focus more on specific key ideas rather than generalized theorems to pinpoint the meaning of the question. Am I correct in saying that all you have to prove for A to be a subgroup is closure and inverses/identity?

8. Jul 29, 2014

gopher_p

First off, your use of the preposition "in" is making it very difficult for me to determine if, no offense, you really understand what you're doing. If you would, please use "is a member/element of", "is a subset of", "is a subgroup of", or the appropriate symbols to make your intended meaning clear so that I can more easily help you.

Also, I realize I'm being kinda hard on you here. Don't take it personally.

That $abH(ab)^{-1}$ is a subset of $G$ is too trivial to mention. That $abH(ab)^{-1}$ is a subgroup of $G$ is not as trivial, but as far as I can tell, it's not really relevant to the problem.

Noting that $(ab)H(ab)^{-1}=(ab)H(b^{-1}a^{-1})=a(bHb^{-1})a^{-1}$ is important, and I think most reasonable people would accept that as true with no further explanation. You should, however, convince yourself why the "obvious" symbol-pushing actually works in this case, though.

Again $bHb^{-1}$ being a subset/subgroup of $G$ is not particularly interesting or relevant to this proof. Furthermore, you don't need to know that to conclude that $b\in G$. $b$ is assumed to be an element of $A$ which is assumed to be a subset of $G$. $b$ is trivially an element of $G$.

Ok ...

Yes, $A$ is a subset of $G$. That's trivial. If by "in" you mean "is a subgroup of", I ask "why?" Why is $A$ a subgroup of $G$, and why was the fact that $aa^{-1}=e$ apparently the key piece needed to finish the proof?

Here is a string of equalities, some of which you've come up with on you own, that I think is important to proving that $A$ is closed under the groups operation of $G$; $$(ab)H(ab)^{-1}=(ab)H(b^{-1}a^{-1})=a(bHb^{-1})a^{-1}=aHa^{-1}=H$$

What I would like for you to do is answer, for yourself, the "why"s. Why is each equality true? Why is this important to showing that $A$ is closed under the groups operation of $G$?

I think that's a fantastic idea. If I'm not mistaken, you're self-studying. In my opinion, there's no point to trying to learn this particular material on your own if you're not aiming for near total understanding.

The are many ways to establish that a subset of a group is a subgroup. The most common for "beginners" are the "one-" and "two-srep" methods. The ""two-step" method involves demonstrating closure of the subset under the group operation and inverses; i.e. showing given $a,b\in A$ that $ab\in A$ (closure under the group operation) and $a^{-1}\in A$ (closure under inverses). The "one-step" method works by showing given $a,b\in A$, that $ab^{-1}\in A$.

9. Jul 29, 2014

Justabeginner

I appreciate you being "hard" on me. It will only make me improve my understanding of the material. Yes, I hope to achieve a thorough understanding so that I may be able to solve the toughest problems. Sorry about the lack of clarity (LaTex) -- I am typing from a mobile phone.

I did not think to make bHb^-1 into H...which was obvious (thank you for pointing this out!). I need to keep practicing proofs so I know how to sieve through pertinent info and flesh out my arguments. That has been my main issue, as is evident in this problem.

Again, thank you for your guidance and explanations.

10. Jul 29, 2014

jbunniii

I will just mention as a side note that the result remains true, with the same proof, if $H$ is any nonempty subset of $G$. It's not necessary to assume that $H$ is a subgroup.

11. Jul 30, 2014

gopher_p

One thing that I found useful when I was first learning to do proofs (and I still do it fairly frequently) is to write on my scratch paper (if you're not using scratch paper, start) all of the things I know/assume and what I am trying to show.

In this case, I am assuming that $a,b\in A$, which means that $aHa^{-1}=H$ and $bHb^{-1}=H$. I am trying to show that $ab\in A$ which means that I am trying to show that $(ab)H(ab)^{-1}=H$. So I 'm trying to demonstrate an equality, which often times involves filling in the gaps between the LHS and the RHS.

Now you actually did the hard work in filling the first part of that gap; recognizing how to rewrite $(ab)H(ab)^{-1}$. I just wrote that out the "right" way - as a string of equalities - for you. The only reason I got the "obvious" second part that you didn't get is that I already had written down what I was gonna do with $aHa^{-1}$ and $bHb^{-1}$ if and when I saw them.

Another way to look at it is to understand that is does not make sense to assume that $a,b\in A$ if I'm not going to somehow use that to my advantage. The only interesting thing that $a,b\in A$ tells me is that $aHa^{-1}=H$ and $bHb^{-1}=H$. So I am expecting at some point to use $aHa^{-1}=H$ and $bHb^{-1}=H$ to say something interesting about $(ab)H(ab)^{-1}$ (and later on $a^{-1}Ha$ when I get around to showing closure under inverses).

12. Jul 31, 2014

Justabeginner

I used your advice to try and piece together the puzzle. I wrote down all the assumptions, givens, and conclusions. I hope this is logical.

For elements a, b in A aHa^-1 = H and bHb^-1=H. Therefore, (ab)H((ab)^-1) = abHb^-1a^-1 = a(bHb^-1)a^-1 = aHa^-1. Now, aHa^-1 is contained within G, therefore A is a subgroup of G.