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The problem is asking for a proof of Hall's theorem for finite solvable groups, namely that there exist Hall $\pi$-subgroups of $G$ a finite solvable group for any set of primes $\pi$ and furthermore that they are all conjugate. I proved existence for all cases, was able to narrow down the potential exceptions of conjugacy to the case where $|G|=p^an$, where $n$ is the order of a Hall $\pi$-subgroup and $p^a$ is the only order possible of a minimal nontrivial normal subgroup $M$. With $N/M$ a minimal normal $q$-subgroup of $G/M$, I can also prove that for $Q∈\text{Syl}_q(N)$, $n$ divides $N_G(Q)$ where $Q$ is assumed nonnormal in $G$ so that by induction all the Hall $\pi$-subgroups of $N_G(Q)$ are conjugate, but how to prove that an arbitrary Hall $\pi$-subgroup is contained in $N_G(Q)$?

Plainly put, I don't understand the justification behind the final sentence of the solution here.

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Let $H$ be an arbitrary Hall $\pi$-subgroup of $G$. It is not necessarily true that $H \le N_G(Q)$, but $|H|=n$ and $H$ is a complement of $M$, so $|H \cap N| = |N/M| = |Q|$. So $H \cap N$ is Sylow $q$-subgroup of $N$ and hence is conjugate to $Q$. That is, there exists $g \in N$ with $(H \cap N)^g = Q$. Now $H \le N_G(H \cap N)$, so $H^g \le N_G(Q)$ and so $H^g$ is a Hall $\pi$-subgroup of $N_G(Q)$ and we are done.

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  • $\begingroup$ Most likely I'm forgetting something evident about the normalizer subgroup of $Q$, but how is that final clause entailed? $\endgroup$ – Feryll Apr 7 '13 at 20:14
  • $\begingroup$ I've added a bit more detail. $\endgroup$ – Derek Holt Apr 7 '13 at 21:53

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