H is a subgroup of its Centralizer iff H is Abelian

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Homework Statement



G is a group, and H is a subgroup of G.
(1) Show H is a subgroup of its Normalizer. Give an example to show that this is NOT necessarily true if H is NOT a subgroup.

(2) Show H is a subgroup of its Centralizer iff H is Abelian

Homework Equations



normalizer NG(H) = {g in G|gHg(^-1)=H}

centralizer CG(H) = {g in G|gh = hg for all h in H}

The Attempt at a Solution



Okay, for part (1) I showed the left coset of H equals to the right coset of H, that is g*H = H*g = H <=> h in H

Can anyone show me an example that it is NOT necessarily true if H is NOT a subgroup? (If H is NOT a subgroup of G, isn't that obvious that it's not necessary a subgroup of its Normalizer?)

For part (2) by definition of the centralizer CG(H)={g in G|gh = hg for all h in H}.
(=>) Suppose H < CG(H), WTS H is Abelian, that is for any elments h1, h2 in H, h1*h2=h2*h1.
so by definition
g*h1 = h1*g => g = h1*g*h1^(-1)
g*h2 = h2*g => g = h2*g*h2^(-1)
So h1*g*h1^(-1) = h2*g*h2^(-1)
Also h1*h2^(-1) is in H by definition of being subgroup of it's centralizer, so g*h1*h2^(-1)=h1*h2^(-1)*g
Okay, I've been trying to drive to h1*h2=h2*h1 but somehow stuck. Can anyone help? I know it's not a complicated problem...

(<=) Suppose H is Abelian, then for any elments h1, h2 in H, h1*h2=h2*h1 => h1*h2^(-1) = h2^(-1)*h1. WTS H < CG(H)
WTS two things:
1) H is nonempty
2) for any h1, h2 in H => h1*h2^(-1) is in H
Obviously H is nonempty since the identity is there.
Let h1, h2 in H
Then since h1*h2=h2*h1 => h1*h2^(-1) = h2^(-1)*h1
How do I show from here that it implies that h1*h2^(-1) is in H?

Thanks!
 

Answers and Replies

  • #2
828
2

Homework Statement



G is a group, and H is a subgroup of G.
(1) Show H is a subgroup of its Normalizer. Give an example to show that this is NOT necessarily true if H is NOT a subgroup.

(2) Show H is a subgroup of its Centralizer iff H is Abelian

Homework Equations



normalizer NG(H) = {g in G|gHg(^-1)=H}

centralizer CG(H) = {g in G|gh = hg for all h in H}

The Attempt at a Solution



Okay, for part (1) I showed the left coset of H equals to the right coset of H, that is g*H = H*g = H <=> h in H

Can anyone show me an example that it is NOT necessarily true if H is NOT a subgroup? (If H is NOT a subgroup of G, isn't that obvious that it's not necessary a subgroup of its Normalizer?)

For part (2) by definition of the centralizer CG(H)={g in G|gh = hg for all h in H}.
(=>) Suppose H < CG(H), WTS H is Abelian, that is for any elments h1, h2 in H, h1*h2=h2*h1.
so by definition
g*h1 = h1*g => g = h1*g*h1^(-1)
g*h2 = h2*g => g = h2*g*h2^(-1)
So h1*g*h1^(-1) = h2*g*h2^(-1)
Also h1*h2^(-1) is in H by definition of being subgroup of it's centralizer, so g*h1*h2^(-1)=h1*h2^(-1)*g
Okay, I've been trying to drive to h1*h2=h2*h1 but somehow stuck. Can anyone help? I know it's not a complicated problem...

(<=) Suppose H is Abelian, then for any elments h1, h2 in H, h1*h2=h2*h1 => h1*h2^(-1) = h2^(-1)*h1. WTS H < CG(H)
WTS two things:
1) H is nonempty
2) for any h1, h2 in H => h1*h2^(-1) is in H
Obviously H is nonempty since the identity is there.
Let h1, h2 in H
Then since h1*h2=h2*h1 => h1*h2^(-1) = h2^(-1)*h1
How do I show from here that it implies that h1*h2^(-1) is in H?

Thanks!

I think I might be a little confused about the second part. Do you mean:
H is a subgroup of G's Centrelizer <=> H is abelian
of
H is a subgroup of H's Centralizer <=> H is abelian ?

In either case, it seems that the => direction (well, both directions) are pretty easy. If H is a subgroup of a centralizer (any centralizer really) then by definition it is abelian. All the stuff in the centralizer commutes with all the elements of the the group of which it is a centralizer, in particular, stuff in the centralizer commute with everything else in the centralizer. This is an abelian group.

The <= direction is not too hard, either, but I'll let you do that one.


BTW: I might have mis-read the question, my apologies if I have.
 
  • #3
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H < CG(H) <=> H is Abelian

CG(H) is the centralizer of H in G.

Being a centralizer of H in G, just saying every element of H commute with every element of G. It does NOT say anything about the relationship between elements inside the H. We need to use the fact that H is a subgroup of G and H is also a subgroup of its centralizer in G to show that for any two elements h1, h2 in H they do commute, that is h1*h2=h2*h1.
 
  • #4
828
2
Ahh, yes, I was thinking of Center; I thought you were talking about the Center of G.

Anyway, as you said: centralizer CG(H) = {g in G|gh = hg for all h in H}

That is, C(H) is all the stuff in G that commutes with all the stuff in H, right? To show => we assume that H is a subgroup of C(H), right? Therefore, if h is in H, then h is also in C and so h commutes with every element of H, right? Doesn't that imply that every element of H is an element of C(H) and therefore commutes with every element of H, making H abelian?
 
  • #5
37
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Yeah, that's true, as simple as it is. Doesn't even need any algebraic proof, just simply explain it out then.
 
  • #6
828
2
Exactly, though I would still call this an algebraic proof. It is just using the definitions of algebraic structures. Of course, you could always do it with set-builder notation, and more symbols if you wanted.
 

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