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Conjugates in the normalizer of a p-Sylow subgroup

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[SOLVED] Conjugates in the normalizer of a p-Sylow subgroup

1. Homework Statement
Let P be a p-Sylow subgroup of G and suppose that a,b lie in Z(P), the center of P, and that a, b are conjugate in G. Prove that they are conjugate in N(P), the normalizer of P (also called stablilizer in other texts I believe).

This is problem 29 from section 2.11 of Herstein's Abstract Algebra, the last problem in Chapter 2.


2. Homework Equations
Any of the standard results in Sylow theory are usable.



3. The Attempt at a Solution
I believe that it would be sufficient to demonstrate that if b = x^(-1)ax then x must lie in the normalizer of P, N(P).

We know that if x is in N(P) then x^(-1)Px = P. We also know that both a and b commute with every element of P. In fact, we also know that x can't be in P, unless a = b. This doesn't seem like it should be too difficult, but I would love a small hint of what direction to take.

Thanks all.
 
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Answers and Replies

139
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*bump*
 
139
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Well, let me put up some more of what I have found - still no solution.

Since a,b [itex] \in Z(P) [/itex] we know that [itex] P \subseteq C(a) [/itex] and [itex] P \subseteq C(b)[/itex]. Since P [itex] \subseteq [/itex] N(P) we also know that [itex] P \subseteq N(P) \cap C(a) [/itex] and [itex] P \subseteq N(P) \cap C(b) [/itex].

Further, [itex]x^{-1} C(a) x [/itex] = [itex] C(b) [/itex] so that the conjugate Sylow subgroup [itex] Q = x^{-1}Px \subseteq C(b)[/itex].

We know that N(P) and N(Q) each have unique Sylow p-subgroups P and Q respectively. If we know in addition that C(a) and C(b) also have unique p-Sylow subgroups then this would solve the problem. However I know of no reason why this must be the case. Anybody see anything obvious that I'm missing (or not obvious :confused:)?

Thanks again all.
 
139
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OK solved, if anyone wants to see solution let me know.
 

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