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I am trying to solve the following exercise: Let $G$ be a finite group and $A$ a Hall abelian subgroup of $G$. Assume that $$ A\le Z(N_G(A)) $$ Then $G$ has a normal $p-$complement for every prime divisor $p$ of $|A|$


I am following the indications given. Argue by induction on $|G|$. If $G=1$, there is nothing to prove. Let $P\in Syl_p(A)$.

  • Assume that $N_G(P)<G$. As $A$ is abelian, $P\triangleleft A$ and so $A\le N_G(P)<G$. Hence $A$ is an abelian Hall subgroup of $N_G(P)$. By hypothesis, $A\le Z(N_G(A))$ and since $N_{N_G(P)}(A)\le N_G(A)$, it follows that $$ A\le Z(N_{N_G(P)}(A)) $$ By the induction hypothesis, it follows that $P$ has a normal complement in $N_G(P)$ i.e. there is $K\triangleleft N_G(P)$ such that $N_G(P)=PK$ and $P\cap K=1$. Since $P$ is abelian, $P=Z(P)$. Moreover, since $P,K\triangleleft N_G(P)$, $[P,K]\le P\cap K=1$ and so $P\le Z(K)$. Hence $P\le Z(N_G(P))$ and, by by Burnside's transfer lemma, $P$ has a normal complement in $G$.

  • Assume that $P\triangleleft G$. If every Sylow subgroup of $A$ is normal in $G$, we are done. Indeed, since $A$ is an Hall subgroup of $G$, every such subgroup is a Sylow subgroup of $G$ and therefore $G$ is nilpotent (and thus $p-$nilpotent for every prime $p$). Therefore assume that there is $Q\le Syl_q(G)$ for $q\neq p$ such that $N_G(Q)<G$. I now show that $P\le Z(N_G(Q))$. By the same reasoning as above, $A\le Z(N_{N_G(Q)}(A))$. Hence $N_G(Q)$ has a normal $p-$complement for every prime divisor $p$ of $|A|$. By the lemma below, $A$ has a normal complement in $N_G(Q)$ i.e. there is $L\le N_G(Q)$ such that $N_G(Q)=AL$ and $A\cap L=1$. Since $A$ is abelian, $P\le Z(A)$ and $P\le C_G(Q)\le N_G(Q)$. Since $P\triangleleft G$, it follows that $P\triangleleft N_G(Q)$. Therefore $[P,L]\le P\cap L\le A\cap L=1$. Hence $P\le Z(L)$ and so $P\le Z(N_G(Q))$.

From there I am stuck.

Above, I have used the following lemma.

Lemma: Let G be a finite group and $A$ an abelian subgroup of $G$. Write $A=P_1...P_n$ where the $P_i$ are the Sylow $p_i-$subgroups of $A$ for $1\leq i\leq n$. Assume each $P_i$ has a normal $(p_i-)$complement $K_i$ in $G$. Then $A$ has a normal complement in $G$ and it is $K_1 \cap...\cap K_n$.

Proof:

We first show that $P_1...P_n \cap \bigcap\limits_{1\leq i\leq n}K_i=1$. Since $A$ is abelian, each $P_i$ is a a normal subgroup of $A$ and so any product of the $P_i$ is a subgroup of $A$ and obviously of $G$. Since $[G:K_1]\wedge |P_2|=1$, it follows that $P_2\le K_1$. Now, by the modular law, $$ P_1P_2\cap K_1=P_2(P_1\cap K_1)=P_2 $$ and thus $$ P_1P_2\cap K_1\cap K_2=P_2\cap K_2=1 $$ The result then follows by induction.

We have $[G:\bigcap\limits_{1\leq i\leq n}K_i] \leq \prod\limits_{1\leq i\leq n} [G:K_i]$. Hence $$ |\bigcap\limits_{1\leq i\leq n}K_i| \geq \dfrac{|G|}{\prod\limits_{1\leq i\leq n} [G:K_i]}=\dfrac{|G|}{\prod\limits_{1\leq i\leq n} |P_i|} $$ Now $|P_1...P_n|=\prod\limits_{1\leq i\leq n} |P_i|$ since the $P_i$ are all coprime. Hence $|P_1...P_n \bigcap\limits_{1\leq i\leq n} K_i| \geq |G|$. And so $G=P_1...P_n \bigcap\limits_{1\leq i\leq n} K_i$.

Last, since $K_i \triangleleft G$ for all i, we have $\bigcap\limits_{1\leq i\leq n}K_i \triangleleft G$ and so $K=\bigcap\limits_{1\leq i\leq n}K_i$ is the normal complement of $P_1...P_n$.


I found the solution.

By the same reasoning as above, $Q\le N_G(Z(Q))$ and so by Burnside's transfer theorem, $Q$ has a normal complement $L$ in $G$. Now as $P\triangleleft G$, $P$ is the unique Sylow $p-$subgroup of $G$ and so $P\le L$. Then $P$ is an abelian normal Hall subgroup of $L$ and so, by Schur E, $P$ has a normal complement $K$ in $L$. Since $P\triangleleft L$, it follows that $L=P\times K$. We have $$G=QL=QKP$$ and we show that $QK$ is a normal complement to $P$ in $G$. Actually we show that $G=P\times QK$. It suffices to show that $[P,QK]=P\cap QK=1$. $Q$ is clearly a $p'-$group. Since $P$ is a Sylow sybgroup of $L$, it follows that $K$ is also a $p'-$group. Since $Q\cap K=1$, $|QK|=|Q||K|$ is a $p'$number. Hence, by Lagrange's theorem, $P\cap QK=1$. Trivially, $[P,Q]=1$. And $[P,K]=1$ since $L=P\times K$. Hence $[P,QK]=1$.


I believe the following proof works.

We show by induction on $|G|$ that $G$ has a normal $p-$complement for every prime divisor $p$ of $|A|$, which gives immediately the result by the lemma. If $G=1$, there is nothing to prove. Let $P\in Syl_p(A)$.

  • Assume that $N_G(P)<G$. As $A$ is abelian, $P\triangleleft A$ and so $A\le N_G(P)<G$. Hence $A$ is an abelian Hall subgroup of $N_G(P)$. By hypothesis, $A\le Z(N_G(A))$ and since $N_{N_G(P)}(A)\le N_G(A)$, it follows that $$ A\le Z(N_{N_G(P)}(A)) $$ By the induction hypothesis, it follows that $P$ has a normal complement in $N_G(P)$ i.e. there is $K\triangleleft N_G(P)$ such that $N_G(P)=PK$ and $P\cap K=1$. Since $P$ is abelian, $[P,P]=1$. Moreover, since $P,K\triangleleft N_G(P)$, $[P,K]\le P\cap K=1$ and so $[P,K]=1$. Hence $P\le Z(N_G(P))$ and, by Burnside's transfer theorem, $P$ has a normal complement in $G$.

  • Assume that $P\triangleleft G$. If every Sylow subgroup of $A$ is normal in $G$, we are done. Indeed, it implies that $A\triangleleft G$ and so by hypothesis, $A\le Z(G)$. By Schur-Zassenhaus, $A$ has a complement $K$ in $G$. This complement is normal in $G$ as trivially $K\le N_G(K)$ and $A \le N_G(K)$ since $A\le Z(G)$ and therefore $G=AK\le N_G(K)$. Therefore assume that there is $Q\le Syl_q(G)$ for $q\neq p$ such that $N_G(Q)<G$.

We show that $P\le Z(N_G(Q))$. By induction, $A$ has a normal complement in $N_G(Q)$ i.e. there is $L\triangleleft N_G(Q)$ such that $N_G(Q)=AL$ and $A\cap L=1$. Since $A$ is abelian, $[P,A]=1$ and $P\le C_G(Q)\le N_G(Q)$. Since $P\triangleleft G$, it follows that $P\triangleleft N_G(Q)$. Therefore $[P,L]\le P\cap L\le A\cap L=1$. Hence $[P,N_G(Q)]=[P,AL]=1$ and so $P\le Z(N_G(Q))$.

We will prove that $P$ is contained in the center of $G$ and since $P$ is normal in $G$, we have that $P\le Z(N_G(P))$, so $P$ has a normal $p$-complement by Burnside's Theorem. If $g \in G$ then $[P,Q^g]=[P^g,Q^g]=[P,G]^g=1$ where the first equality follows from the fact that $P$ is normal in $G$ and the last from the fact that $[P,Q]=1$ since $P<Z(N_G(Q))$. This means that $P$ commutes with all the conjugates of $Q$ in $G$ and then commutes with the normal closure $Q^G$. If $Q^G=G$ then $P$ commutes with the whole $G$ and it is in the center, so we are done. If $Q^G<G$ then $Q^G$ is a normal subgroup of $G$ containing $Q$ that is a $q$-Sylow of $G$ and by Frattini Argument $G=N_G(Q)Q^G$. But $P$ commutes with both $N_G(Q)$ and $Q^G$ and then commutes with their product $G$, in other words $P<Z(G)$.

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    $\begingroup$ Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Shahab Mar 21 '16 at 15:56
  • $\begingroup$ If a sylow p-subgroup is normal then by Schur-Zassenhaus theorem it has an extension that splits, nevertheless this complement does not need to be normal. Indeed, for G the symmetric group of order 6, a cyclic group of order 3 is a Sylow normal subgroup and cannot exists a normal complement or order 2, being G non abelian. $\endgroup$ – Lorban Jan 4 '17 at 10:57
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I propose an alternative solution for the last part of the proof. Summarizing, suppose $G$ finite group and $H$ is a Hall subgroup that is central in its normalizer, we have to prove that for each prime $p$ that divides $|H|$, $G$ has a normal $p$-complement. We work by induction on $|G|$. If $P \in Syl_p(H)$ you have already proved that $G$ has a normal $p$-complement in the case where $P$ is not normal in $G$ and also in the case where every Sylow subgroup of $H$ (that are indeed Sylow's in $G$) is normal in $G$. So you have assumed $P$ is a normal $p$-Sylow of $G$ contained in $H$ and there exists for a prime $q$, different from $p$, a $q$-Sylow $Q$ of $H$ that is not normal in $G$, i.e. $N_G(Q)<G$. As you have shown, we know that $P$ centralizes $N_G(Q)$. We will prove that $P$ is contained in the center of $G$ and since $P$ is normal in $G$, we have that $P\le Z(N_G(P))$, so $P$ has a normal $p$-complement by Burnside's Theorem. If $g \in G$ then $[P,Q^g]=[P^g,Q^g]=[P,G]^g=1$ where the first equality follows from the fact that $P$ is normal in $G$ and the last from the fact that $[P,Q]=1$ since $P<Z(N_G(Q))$. This means that $P$ commutes with all the conjugates of $Q$ in $G$ and then commutes with the normal closure $Q^G$. If $Q^G=G$ then $P$ commutes with the whole $G$ and it is in the center, so we are done. If $Q^G<G$ then $Q^G$ is a normal subgroup of $G$ containing $Q$ that is a $q$-Sylow of $G$ and by Frattini Argument $G=N_G(Q)Q^G$. But $P$ commutes with both $N_G(Q)$ and $Q^G$ and then commutes with their product $G$, in other words $P<Z(G)$. This completes the proof.

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    $\begingroup$ Thanks very much Lorban. It looks great to me. i have included it in a revised proof above. $\endgroup$ – Nicolas Jan 3 '18 at 10:41

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