# Conjugate Subgroups of a Finite Group

1. Jul 30, 2014

### Justabeginner

1. The problem statement, all variables and given/known data
Two subgroups of G, H and K are conjugate if an element a in G exists such that aHa^-1= {aha^-1|elements h in H}= K

Prove that if G is finite, then the number of subgroups conjugate to H equals |G|/|A|.

2. Relevant equations
A={elements a in G|aHa^-1=H}

3. The attempt at a solution

Stabilizer of H also known as NG(H) is {a: aHa^-1 = H}.
By Lagrange's Theorem, number of left cosets in H = |G|/|H|.
NG (H) is a normalizer and NG (H) = {elements g in G| gHg^-1 = H}.
A is normalizer and A= NG (H) because A={elements a in G|aHa^-1=H}.
Therefore, no. of subgroups conjugate to H, is |G|/|A|.

This argument is terribly weak and I have no idea how to formalize the proof. I want to just use the concept of subgroup/group properties not normalizers/stabilizers/centralizers because I clearly do not understand the latter well enough to do so.

I appreciate any help. Thank you.

2. Jul 30, 2014

### jbunniii

Try defining a mapping from $G/A$ (the set of cosets of $A$) to the set of conjugates of $H$. The natural thing to try is to map the coset $xA$ to the conjugate $xHx^{-1}$. Can you show that this is a well-defined mapping, i.e., if $xA = yA$ then $xHx^{-1} = yHy^{-1}$? Is the mapping a bijection?

3. Jul 30, 2014

### Justabeginner

ψ: G/A -> H, where G/A = {elements x, y in A| xA=Ax and yA=Ay} and H = {elements x, y in H|xHx^-1 and yHy^-1 = H}
I do not think the mapping is defined correctly.

4. Jul 30, 2014

### jbunniii

No, you've created new definitions for G/A and H, but they are already defined!

G/A is the set of (left) cosets of A. In other words, $G/A = \{xA | x \in G\}$.

$H$ is a subgroup of $G$ as given in the problem statement. Let's define $X = \{xHx^{-1} | x \in G\}$, the set of conjugates of $H$.

My suggestion is to define $\psi : G/A \rightarrow X$ by $\psi(xA) = xHx^{-1}$.

If you can show that $\psi$ is a well-defined bijection, then we can conclude that the size of $G/A$ must be equal to the size of $X$, which is the result you need.

So you need to show:
1. $\psi$ is well defined, i.e. if $xA = yA$ then $xHx^{-1} = yHy^{-1}$
2. $\psi$ is a bijection

5. Jul 30, 2014

### Justabeginner

1. xA= yA if x^-1y is in H
A = x^-1y

Similarly, Ax^-1y= Ay^-1 if x^-1y is in H. Again, A= x^-1y.
Then, the mapping is defined.

2. (xA)^-1 = (yA)^-1
(xA) (xA)^-1 = (xA) ((yA)^-1) e = (xA) ((yA)^-1)
e(yA) = (xA)((yA)^-1)(yA)= yA = xA
So, (xA)^-1 = (yA)^-1 <---> yA = xA
The homomorphism is bijective.

Therefore, if xA has order n in G/A, then xHx^-1 in X also has order n.
Since G/A and X have the same order, and X is the set of conjugates on H, then |X| = |G|/|A|.

Is this accurate?

Last edited: Jul 30, 2014
6. Jul 30, 2014

### jbunniii

You mean $x^{-1}y \in A$, not $H$. Also, this is an "if and only if", which will be useful later.
No, this makes no sense. The left hand side is a set of elements (indeed, a subgroup of $G$), whereas the right hand side is a single element.

You mean "is in $A$".
Again, this makes no sense.

The goal is to show that if $xA = yA$, then $xHx^{-1} = yHy^{-1}$. So assume $xA = yA$. Thus $x$ and $y$ are in the same coset of $A$. Therefore $y^{-1}x \in A$. By definition of $A$, this means that $(y^{-1}x)H(y^{-1}x)^{-1} = H$. Can you finish the proof from here?

I have no idea what you're doing here. You need to show that $\psi$ is a bijection. (By the way, it won't be a homomorphism.) To show that $\psi$ is a bijection, you need to show that it is an injection and a surjection. Let's show that it is an injection. To do this, we need to show that if $xHx^{-1}= yHy^{-1}$ then $xA = yA$. So suppose that $xHx^{-1} = yHy^{-1}$. Multiplying on the left by $y^{-1}$ and on the right by $y$, we get $y^{-1}xHx^{-1}y = H$. Can you finish the proof?

Last edited: Jul 30, 2014
7. Jul 30, 2014

### Justabeginner

If xy^-1Hyx^-1 = H, and y^-1Hy = H then x(y^-1Hy)x^-1 = xHx^-1. So, when xHx^-1 = H = H, the mapping of H (onto itself) is well defined.

If y^-1xHx^-1y = H, and H = xHx^-1 , then y^-1Hy = H and H = H?

8. Jul 30, 2014

### jbunniii

No, you cannot assume that $y^{-1}Hy = H$. It won't be true in general. If you go back to what I wrote above, I showed that if $xA = yA$, then $(y^{-1}x)H(y^{-1}x)^{-1} = H$. The goal is to show that this implies that $xHx^{-1} = yHy^{-1}$. So you need to show how to go from
$$(y^{-1}x)H(y^{-1}x)^{-1} = H$$
to
$$xHx^{-1} = yHy^{-1}$$
Same problem here. You can't assume that $H = xHx^{-1}$. It is not true in general. You need to show that if $y^{-1}xHx^{-1}y = H$, then $xA = yA$. I'll do another step for you. Given that $y^{-1}xHx^{-1}y = H$, we can rewrite this slightly as
$$y^{-1}xH(y^{-1}x)^{-1} = H$$
Therefore, $y^{-1}x \in A$. (Why?)

9. Jul 30, 2014

### Justabeginner

Then (y^-1x)H(x^-1y) = H and xHx^-1 = yHy^-1 by multiplication.

y^-1x is contained in A because if x and y are in the same coset of A, and xy^-1 is contained in A, then y^-1x should also be contained in A. (Is x ~ y an equivalence relation in A?)

10. Jul 30, 2014

### jbunniii

Looks good.
No, this is arguing circularly. You need to explain why
$$y^{-1}xH(y^{-1}x)^{-1} = H$$
implies
$$y^{-1}x \in A$$
What is the definition of $A$?

11. Jul 30, 2014

### Justabeginner

A={elements a in G|aHa^-1=H}

I do not understand if we cannot use the assumption that xHx^-1 = H then how can we come to the conclusion that y^-1x is in A?

12. Jul 30, 2014

### jbunniii

Look again at $y^{-1}xH(y^{-1}x)^{-1} = H$. If I define $a = y^{-1}x$, then this equation becomes
$$aHa^{-1} = H$$
Therefore...? What can you say about $a$?

13. Jul 30, 2014

### Justabeginner

Then element a is contained in A (and H).

14. Jul 30, 2014

### jbunniii

Yes. So what can you conclude?
Not necessarily!

15. Jul 30, 2014

### Justabeginner

Oh I realize my mistake. H is a subgroup of A?

If a is an element in A, then y^-1x is also contained in A. Then xA = yA?

16. Jul 30, 2014

### jbunniii

Yes, that's correct. The following relation holds: $H \unlhd A \leq G$ (in words, $H$ is a normal subgroup of $A$, and $A$ is a subgroup of $G$). Incidentally, $A$ is called the normalizer of $H$, because it is the largest subgroup of $G$ in which $H$ is normal.
Right, $a = y^{-1}x$, so $y^{-1}x \in A$. Therefore, $x \in yA$, in other words, $x$ and $y$ are in the same coset of $A$, in other words, $xA = yA$.

17. Jul 30, 2014

### Justabeginner

And now because ψ is well defined and isomorphic, we determine that |G|/|A| = |X|.

18. Jul 30, 2014

### jbunniii

Almost correct. In fact, $\psi$ is a bijection but it is not necessarily a homomorphism. (Indeed, $G/A$ is not even a group unless $A$ happens to be normal in $G$.) But if you change one word, it's correct:

And now because ψ is well defined and bijective, we determine that |G|/|A| = |X|.

(Here we have used the fact that $|G/A| = |G|/|A|$ by Lagrange's theorem.)

By the way, you have shown that $\psi$ is an injection, but you haven't yet shown that is a surjection. Fortunately, that part is easy. Can you see why it's true?

19. Jul 31, 2014

### Justabeginner

For every xA in G/A, there must be an xHx^-1 in X. This is explained by the mapping of Psi(xA) onto xHx^-1, therefore the mapping is surjective as well?

20. Jul 31, 2014

### jbunniii

The goal is to show that $\psi : G/A \rightarrow X$ is a surjection. Therefore your proof should start with "for every $xHx^{-1} \in X$, the element ____ of $G/A$ satisfies $\psi($___$) = xHx^{-1}$." (Fill in the blank.)