# Normalizer N_G(X) equal to another set when o(X)<infty

1. Sep 9, 2015

### jackmell

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

Prove that if a set $X\subseteq G$ is finite then it's Normalizer in G, $N_G(X)$ is also equal to the set: $A=\{g\in G : gXg^{-1}\in X\}$

2. Relevant equations

Given $X\subseteq G$, then the normalizer of X in G is defined as:
$N_X(G)=\{g\in G: gXg^{-1}=X\}$

3. The attempt at a solution

First I have to note the subtle difference in the two sets: The normalizer set is equal to the set of g such that the conjugate $\sigma_g(X)$ is equal to X whereas the set $A$ above is when the conjugate $\sigma_g(X)$ only maps to a subset of X. I then need to show when $X$ is finite, these two sets are equal. That is, when $o(X)<\infty$ then I would have:

$\{g\in G:gXg^{-1}=X\}=\{g\in G: gXg^{-1}\subseteq X\}$

I believe $N_G(X)\subseteq A$ since the set of $g$ which map to all of $X$ by conjugation is certainly going to be in the set of g mapping to a subset of X. So that if I can show that likewise $A\subseteq N_G(X)$ then surely $N_G(X)=A$.

I would then need to prove that if $g\in A$, in which $\sigma_g(X)$ is mapping a subset of X to itself, then it must necessarily map all of $X$ to itself. Then I would think that includes the possibility of showing that if for $g\in A$, such that $\sigma_g(x)\in X$ for a single $x\in X$ then necessarily, $\sigma_g$ must map all of $X$ to itself.

Now, I can show that if $\sigma_g(x_1)=x_2\in X$ then $\sigma_g$ maps also $x_1^{-1}, x_2, x_2^{-1}, x_1 x_2$ into $X$. However I cannot show that if $\sigma$ is mapping a single element into $X$, then it must map all of $X$ into itself.

Jack

2. Sep 9, 2015

### wabbit

Almost. You need to prove that if $gXg^{-1}\subseteq X$ then $gXg^{-1}=X$. Consider the map $\sigma_g:X\rightarrow X$. Is it injective? Surjective?

3. Sep 9, 2015

### jackmell

Ok thanks wabbit. First, $\sigma$ is an automorphism so it's bijective but as I understand it, $\sigma$ is the bijection:
$\sigma: G\to G$.

However, that doesn't mean that $\sigma$ would map $X$ to itself. So the map $\sigma(X)$ is injective since $\sigma_g(G)$ is bijective over $G$ but I don't see how I could say $\sigma_g(X)$ is also surjective in $X$, that is, $\sigma$ could still be a bijection over $G$ but not bijective over a subset of $G$ or do I have that wrong?

Edit:
Wait, I may have something:

Let $gXg^{-1}\subseteq X$. then $\;\forall\; x\in X, gXg^{-1}\in X$. But $\sigma$ is one-to-one so that $gXg^{-1}$ must map to all of $X$.

Last edited: Sep 9, 2015
4. Sep 9, 2015

### wabbit

Right, I should have been more specific, the map to consider is $\sigma_g^X : X\rightarrow X, x\rightarrow gxg^{-1}$ which exists because of the assumption $gXg^{-1}\subseteq X$. Finiteness of X is key of course.

5. Sep 9, 2015

### jackmell

Thanks a bunch wabbit. That's actually part 2 of the problem (due Tomorrow) but I don't wish to abuse the privilege of you guys helping me so will try and do it by myself without help. Sides, I'll have to cite you in my homework assignment for part 1: "Received help from PF homework helpers see: <such and such> and it might look bad if I got help with both parts. :)

I'll work on it.

Jack

6. Sep 9, 2015

### wabbit

Not sure what part 2 is, but the fact that X is finite is necessary to prove statement 1, so make sure you use it explicitly : )

7. Sep 9, 2015

### jackmell

Ok then can you help me understand why? That's not directly related to the second part. Is it because we can't equate infinite sets? For example the reals and rationals which are both infinite with $\mathbb{Q}\subset \mathbb{R}$ but different cardinality so if we were dealing with infinite sets I could not make the statement that the set $\{g\in G: gXg^{-1}\subset X\}=X$ if they were both infinite?

8. Sep 9, 2015

### wabbit

Close - check again your statement
In other words you're saying $\sigma_g^X$ is one-to-one hence it must be onto. You need to explain why this inference is true.

9. Sep 9, 2015

### jackmell

Ok, since $X\subseteq G$ and is finite, then $|X|\leq |G|$. And thus since $\sigma_g(X)$ is one-to-one, then it must map onto all of $X\;\forall\; x\in X$. I think then that's where it's important that the size of $X$ is finite because if it were infinite, then I don't this this part would be valid.

10. Sep 9, 2015

### wabbit

Yes. In fact a set S is finite if and only if any injective map from S to itself is bijective.

11. Sep 11, 2015

### jackmell

Hi wabbit,

If you're still reading this, would you mind looking at the second part of this problem since I've already tried working it myself and have turned it in? I posted a question regarding infinite non-abelian groups in the Linear and Abstract Algebra sub-forum yesterday to help me use a different group for the solution than what my professor suggested using and he did go through a possible solution quickly in class but I still don't think I got it right.

https://www.physicsforums.com/threa...belian-groups-not-gl_n-g.831679/#post-5224609

Ok thanks,
Jack