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This question is from the Proposition A.23.2 in page 520 of the book ``Berkovich, Yakov, and Zvonimir Janko. Berkovich, Yakov; Janko, Zvonimir: Groups of Prime Power Order. Vol. 2. Walter de Gruyter, 2008.''

Suppose $G$ is a finite nontrivial group.

$F$ is a centralizer in $G$ is means that $F$ is a centralizer of an element.

$F$ is free means that $F\neq G$ and if a centralizer $C_G(a)\leq F$, then either $C_G(a)={1}$ or $C_G(a)=F$ and if a centralizer $C_G(b)\neq G$, $F\leq C_G(b)$, then $C_G(b)=F$.

The author first prove that $F=C_G(x)$, where $x\in G$ and the order $|x|$ of $x$ is a power of a prime $p$.

And he prove that if $y\in F$ and $|y|=q^r$, for some positive integer $r$ and prime $q\neq p$, then $y\in Z(F)$, the center of $F$.

I already know the reason of these two results, but the author then says that $F=P\times A$, where $P$ is a Sylow $p$-subgroup of $F$ and $A$ is abelian (Use Burnside's normal $p$-complement theorem),

I don't know how to use the Burnside's normal $p$-complement theorem in here.

Could anyone give me a favor, thanks very much.

bfhaha
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1 Answers1

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The statements you do believe imply that every element of order prime to $p$ in $F$ is in $Z(F).$ Actually, I think I agree with you. The version of Burnside's normal $p$-complement theore that is usually stated is that if $P \in {\rm Syl}_{p}(G)$ and $P \leq Z(N_{G}(P)),$ then $G$ has a normal $p$-complement, but this requires that $P$ should be Abelian, and I don't think that is necessarily true in this situation. There are ways around it: you can use Frobenius's normal $p$-complement theorem,, since $N_{F}(Q)/C_{F}(Q)$ is certainly a $p$-group for each $p$-subgroup $Q$ of $F.$ Then the normal $p$-complement theorem of Frobenius says that $F$ has a normal $p$-complement $K$ and $F = PK$ for $P \in {\rm Syl}_{p}(F).$ But we already have seen that $K \leq Z(F)$ so $K$ ia Abelian and $F = P \times K.$ You can also use the Schur-Zassenhaus theorem, since $P \lhd F,$ as every element of order prime to $p$ in $F$ is central. Hence $P$ is complemented in $F$ and that complement is contained in $Z(F)$ as before.

Later edit: I suppose that there are ways to do this via Burnside's normal $p$-complement theorem. Another way without Burnside's theorem is to take $A \lhd P$ be be a maximal Abelian normal subgroup of $P.$ Then $C_{P}(A) = A,$ and $C_{P}(A)$ is a Sylow $p$-subgroup of $C_{F}(A).$ Hence $C_{F}(A)$ is Abelian, and if $K$ is the direct product of its Sylow $q$-subgroups for all primes $q \neq p,$ then $K$ char $C_{F}(A) \lhd F,$ so $K \lhd F$ is a normal $p$-complement for $F.$ But one way to use Burnside's theorem is to say that for each prime $q \neq p,$ a Sylow $q$-subgroup $Q$ of $F$ is in $Z(N_{F}(Q)),$ so $F$ has a normal $q$-complement, and then proceed by induction on the number of prime divisors of $|F|$.

Geoff Robinson
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  • Dear Robinson, thank you for your answer, you give me a new direction, but I have a question.

    How do we know that $P\lhd F$? We need to show that for any $g\in F$, $gPg^{-1}\subseteq P$. If $\gcd{(|g|, p)}=1$, then $g\in Z(F)$ and $gPg^{-1}=P$. But if $\gcd{(|g|, p)}=p$, does it still must be $gPg^{-1}\subseteq P$?

     – bfhaha May 31 '14 at 04:31
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    We know that $P \lhd F$ for the following reason: note that $P \leq N_{F}(P)$ clearly. For every prime $q \neq p,$ and $Q \in {\rm Syl}{q}(F),$ we have $Q \leq Z(F) \leq N{F}(P).$ Hence $N_{F}(P)$ contains a Sylow $r$-subgroup of $F$ for each prime divisor $r$ of $|F|$, so that $|N_{F}(P)|$ is divisible by $|F|$ and we must have $N_{F}(P) = F.$ – Geoff Robinson May 31 '14 at 07:47
  • How excellent the answer is, how could you know such method. Thank you so much. – bfhaha May 31 '14 at 08:25