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I am trying to give a proof of the conjugacy of the Sylow system.

One of the alternative descriptions of the Hall's theorem is that

$G$ is solvable if and only if G has a Hall $p_i'$-subgroup $H_i$ for every prime divisor of $|G|$. ($|G|=\prod_{i=1}^n p_i^{a_i} $)

Now suppose that $G$ is solvable. As $H_1 \cap H_2 \dots H_{i-1} \cap H_{i+1}\dots \cap H_n = P_i$ is a Sylow $p_i$-subgroup of $G$, we can then obtain a Sylow system $S=\{ P_1,P_2 \dots P_n \}$ generated by $\{ H_1, H_2\dots H_n \}$. We have already known that all Hall $\pi$-subgroups are conjugate, hence we have another Sylow system $\{ Q_1,Q_2 \dots Q_n \}$ generated by $\{ {H_1}^{x_1}, {H_2}^{x_2}\dots {H_n}^{x_n} \}$. And I want to prove that these two Sylow systems are conjugate.

My attempt:

First, we can simplify this problem by assuming that $\{ Q_1,Q_2 \dots Q_n \}$ generated by $\{ {H_1}^{x_1}, {H_2}\dots {H_n} \}$ for some $x_1 \in G$. (In fact, if the proof is complete under this circumstance, then we can prove that Sylow systems generated by $\{ {H_1}^{x_1}, {H_2}\dots {H_n} \}$ and $\{ {H_1}^{x_1}, {H_2}^{x_2}, {H_3}\dots {H_n} \}$ are conjugate. Then the proof is complete by repeating this progress.)

I want to prove this by induction on $n$. By the argument above, we have $P_1=Q_1$. Since $G=P_1P_2\dots P_n=Q_1Q_2\dots Q_n$, we have $H=P_2P_3\dots P_n=Q_2Q_3\dots Q_n \le G$. By induction, $(P_2, P_3 \dots P_n)=({Q_2}^x, {Q_3}^x \dots {Q_n}^x)$ for some $x \in H$. So it suffices to show that $x \in N_G(P_1)$. However, I failed to do it. And I'm starting to think that $x$ is not necessarily belongs to $N_G(P_1)$. Can anyone enlighten me with what I'm missing? Or my way isn't gonna work? Then how to prove this? Thanks in advance.

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I think it might be better to use induction on $|G|$. Let $N$ be a minimal normal subgroup of $G$. Then $N$ is an elementary abelian $p$-group for some prime $p$, and we may assume that $p=p_1$.

By induction, the Sylow systems are conjugate in $G/N$. So by replacing $Q_i$ by $Q_i^g$ for the conjugating element $g$, we can assume that $Q_iN=P_iN$ for all $i$, which means that $P_1=Q_1$, since $N \le P_1 \cap Q_1$.

Now $P_2 \cdots P_n$ and $Q_2 \cdots Q_n$ are Hall subgroups of $P_2 \cdots P_nN$, so they are conjugate in $P_1 \cdots P_nN$, and in fact they are conjugate by an element of $N$ (alternatively, that follows from the Schur-Zassenhaus Theorem). Conjugating by an element of $N$ does not change $P_1$, so we can now assume in addition that $P_2 \cdots P_n = Q_2 \cdots Q_n$, and we still have $P_1=Q_1$.

Now $P_2 \cdots P_n = Q_2 \cdots Q_n$ together with $P_iN=Q_iN$ implies $P_i = Q_i$ for $2 \le i \le n$, so we are done.

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  • $\begingroup$ Thanks a lot! I have tried this way by considering $G/O_p(G)$ for some prime divisor of $|G|$. And I stuck where $P_2P_3...P_n O_p(G)=Q_2Q_3...Q_n O_p(G)$. Enlightened by your answer, I realized that I stuck there because I forgot $O_p(G) \le P_1 \cap Q_1$. $\endgroup$ – Johnny May 8 '17 at 17:21
  • $\begingroup$ A reason why I gave up using induction on $|G|$ was that I thought some information of Hall $p'$-subgroups should be used and is hard to use it in this proof. So is there any relevance between the conjugacy of Hall $p'$-subgroups and the conjugacy of the Sylow system? Like the elements satisfy some conditions? $\endgroup$ – Johnny May 8 '17 at 17:31
  • $\begingroup$ Sorry this isn;t really an area I know much about, so I don''t have any deeper insight into what is going on. But (as you probably noticed) the argument in my answer would work equally well with $O_p(G)$ in place of $N$. $\endgroup$ – Derek Holt May 8 '17 at 19:18

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