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suppose that $G$ is finite group and $P$ is aabelian $p$-sylow subgroup of $G$ and $H=N_{G}(P)$.

show that $P$ is normal complement in $G$ if and only if $H$ has normal subgroup $Q$ which $H=P\times Q$.

my solution:suppose $K$ is normal complement of $P$ in $G$ ,put $Q=K \cap H$ ,

for other side suppose $H=P \times Q$ and $Q \triangleleft H$ ,

now please check that the continue is right or wrong:

because $P \triangleleft H$ , $Q \triangleleft H$ we have $[P,Q] \leq P $ and $[P,Q] \leq Q$,so $[P,Q]\leq P\cap Q=1$ ,therefore $P \leq C(Q)$ and then $P \leq Z(H)$ and by Burnside's Theorem we have $P$ is normal complement in $G$ .

thanks a lot.

user1729
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kpax
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1 Answers1

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In principle your solution is correct, but it can be somewhat streamlined. Let us repeat the statement.

Proposition Let $G$ be a finite group and $P \in Syl_p(G)$ be abelian. Then the following are equivalent.

(a) $P$ has a normal $p$-complement in $G$.
(b) $N_G(P)=P \times Q $, with $Q \unlhd N_G(P)$.

Proof (a) $\Rightarrow$ (b). Assume $G=PN$, with $N \unlhd G$ and $P \cap N=1$. Put $Q=N \cap N_G(P)$, which is normal in $N_G(P)$. By Dedekind's Modular Law one has $N_G(P)=PN \cap N_G(P)=P(N \cap N_G(P))=PQ.$ Clearly $P \cap Q \subseteq P \cap N=1$. Since $P$ and $Q$ are both normal in $N_G(P)$, $N_G(P)$ is the internal direct product of $P$ and $Q$.
(b) $\Rightarrow$ (a). Let $N_G(P)=P \times Q$, with $Q \unlhd N_G(P)$. Since $P$ is abelian, $Z(N_G(P))=Z(P \times Q)= P \times Z(Q)$. hence $P \subseteq Z(N_G(P))$ and one can apply Burnside's Theorem to conclude that $P$ has a normal $p$-complement.

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Nicky Hekster
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