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

  • Thread starter jackmell
  • Start date
  • Tags
    Set
In summary: Close - check again your statement The Attempt at a Solution [/B]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
  • #1
jackmell
1,807
54
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\}##

Homework Equations



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

The Attempt at a Solution


[/B]
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.

Ok thanks for reading,
Jack
 
Physics news on Phys.org
  • #2
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?
 
  • Like
Likes jackmell
  • #3
wabbit said:
Almost. You need to prove that if ##gXg^{-1}\subset X## then ##gXg^{-1}=X##. Consider the map ##\sigma_g:X\rightarrow X##. Is it injective? Surjective?

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:
  • #4
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.
 
  • Like
Likes jackmell
  • #5
wabbit said:
Finiteness of X is key of course.

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
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
wabbit said:
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 : )

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
Close - check again your statement
jackmell said:
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##.
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
wabbit said:
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.

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
Yes. In fact a set S is finite if and only if any injective map from S to itself is bijective.
 
  • Like
Likes jackmell
  • #11
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
 

1. What is the definition of Normalizer N_G(X)?

The normalizer N_G(X) of a subset X in a group G is the largest subgroup of G that contains X as a normal subgroup. In other words, it is the set of elements in G that commute with every element in X.

2. How is Normalizer N_G(X) related to other subgroup concepts?

The normalizer N_G(X) is closely related to the centralizer C_G(X), which is the set of elements in G that commute with every element in X. The normalizer is also a subgroup of the stabilizer Stab_G(X), which is the set of elements in G that fix every element in X under a given action.

3. Can Normalizer N_G(X) be equal to another set if o(X) is infinite?

No, the normalizer N_G(X) can only be equal to another set if the order of X (o(X)) is finite. If o(X) is infinite, then N_G(X) will always be a proper subgroup of G.

4. How is Normalizer N_G(X) calculated in practice?

In practice, the normalizer N_G(X) can be calculated by finding the centralizer of X and then taking the normal closure of that subgroup. The normal closure is the smallest normal subgroup that contains the centralizer.

5. Why is the concept of Normalizer N_G(X) important in group theory?

The normalizer N_G(X) is important in group theory because it helps us understand the structure of a group and its subgroups. It also has applications in other areas of mathematics, such as algebraic geometry and number theory.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
983
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
455
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
343
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
462
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
942
Back
Top