1. ## Centralizer Proof Attempt

H be a subgroup of G. Show C(C(C(H)))) = C(H)

hint: show if A is a subgroup of B then C(A) is a subgroup of C(B).

I started off trying to prove the hint.

If A is a subgroup of B then I let there intersection be K and then C(K) has only elements of C(A) but they are also contained in C(B) therefor C(A) is a subgropu of C(B).

Not sure if this is correct and if it is not sure how to apply this to the main question.

Any help greatly apreciated

2. Originally Posted by Niall101
H be a subgroup of G. Show C(C(C(H)))) = C(H)

hint: show if A is a subgroup of B then C(A) is a subgroup of C(B).

I started off trying to prove the hint.

If A is a subgroup of B then I let there intersection be K and then C(K) has only elements of C(A) but they are also contained in C(B) therefor C(A) is a subgropu of C(B).

Not sure if this is correct and if it is not sure how to apply this to the main question.

Any help greatly apreciated
Hint: I guess this should work

Let $\displaystyle x \in C(H)$ i.e.$\displaystyle xh=hx$ for all $\displaystyle h \in H$

Consider any $\displaystyle y \in C(C(H))$. Then $\displaystyle yc=cy$ for all $\displaystyle c \in C(H)$. Specifically $\displaystyle yx=xy$ for any $\displaystyle y \in C(C(H))$. Thus $\displaystyle x \in C(C(C(H)))$

I think the reverse will work out similarly.

3. Originally Posted by Niall101
H be a subgroup of G. Show C(C(C(H)))) = C(H)

hint: show if A is a subgroup of B then C(A) is a subgroup of C(B).

I started off trying to prove the hint.

If A is a subgroup of B then I let there intersection be K and then C(K) has only elements of C(A) but they are also contained in C(B) therefor C(A) is a subgropu of C(B).

Not sure if this is correct and if it is not sure how to apply this to the main question.

Any help greatly apreciated

If by C(A) you mean the centralizer in G of the subgroup A then it is false that $\displaystyle A\subset B \Longrightarrow C(A)\subset C(B)$, and a simple example: $\displaystyle In\,\,S_3\,,\,\,<(12)>\, \leq S_3\,,\,\,but\,\, C(<(12)>)=<(12)>\nsubseteq \{(1)\}=C(S_3)$

Tonio

4. thanks! so basically the hint given is not needed?

would this result hold for the normalizer of H?

5. Originally Posted by Niall101
thanks! so basically the hint given is not needed?

Much worse: the hint given is completely false.

would this result hold for the normalizer of H?

Nop, that's also false.

Tonio

6. Originally Posted by Niall101
thanks! so basically the hint given is not needed?

Much worse: the hint given is completely false.

would this result hold for the normalizer of H?

Nop, that's also false.

Tonio

7. Originally Posted by Niall101
thanks! so basically the hint given is not needed?

would this result hold for the normalizer of H?
In your original problem, after Tonio's quote, I am trying to prove the reverse (without using the hint). But have not been able to do so as yet.

8. Originally Posted by Niall101
H be a subgroup of G. Show C(C(C(H)))) = C(H)

hint: show if A is a subgroup of B then C(A) is a subgroup of C(B).

I started off trying to prove the hint.

If A is a subgroup of B then I let there intersection be K and then C(K) has only elements of C(A) but they are also contained in C(B) therefor C(B) is a subgropu of C(A).

Not sure if this is correct and if it is not sure how to apply this to the main question.

Any help greatly apreciated
My Mistake on the first post

9. Originally Posted by aman_cc
In your original problem, after Tonio's quote, I am trying to prove the reverse (without using the hint). But have not been able to do so as yet.
Is it correct to say the following:
If $\displaystyle y \in C(C(H))$ => $\displaystyle y \in C(H)$?

Here is the argument -
$\displaystyle yz=zy$ for all $\displaystyle z \in C(H)$
$\displaystyle z^{-1}h=hz^{-1}$ for all $\displaystyle h \in H$
Thus, $\displaystyle yh=hy$ $\displaystyle h \in H$

Sorry - Please ignore - This is wrong !

10. Originally Posted by Niall101
My Mistake on the first post

It still makes no sense, and the hint remains and remains false: if A is a sbgp. of B, then their interesection is A itself...not very useful, is it?

Tonio