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Let $G$ be a finite group, $N$ normal subgroup with index in $G$ divisible by $p$ prime and suppose that a Sylow $p$-subgroup of $G$ is cyclic. Then $N$ has a normal $p$-complement. This is the exercise 5C.12 of "Finite Group Theory", Isaacs. Can someone give any hint? Now I don't have any idea.

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Let $P \in {\rm Syl}_p(G)$, $Q =N \cap P$. To apply Burnside's Transfer Theorem, you need to show that $Q$ is central in $N_N(Q)$. We may as well assume that $N = N_N(Q)$ and $G = PN$. Then $N=QH$ where $H$ is a $p$-complement of $Q$ in $N$.

Let $P = \langle g \rangle$. Then $H^g$ is another $p$-complement, so it is conjugate to $H$ by an element $h$ of $Q$. Then by replacing $g$ by $gh^{-1}$ (which also generates $P$), we get $H^g = H$. Let $p^k$ be the highest power of $p$ that divides $|G/N|$. Then $g^{p^k}$ generates $Q$and normalizes $H$ and hence $[H,Q] \le Q \cap H = 1$, and we are done.

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    $\begingroup$ Note small correction at the end. I had $g^p$ before but it should be $g^{p^k}$ as in the corrected version. $\endgroup$ – Derek Holt Jan 5 '17 at 9:52
  • $\begingroup$ Yes I saw it yesterday night, not a big deal to fix. Again, thank you! $\endgroup$ – Lorban Jan 5 '17 at 10:19
  • $\begingroup$ @DerekHolt, can you explain me please, why we can assume without loss of generality that $N=N_N(Q)$ and $G=PN$. I understand that $G=NN_G(Q)$. Thank you. $\endgroup$ – Mikhail Goltvanitsa Mar 31 '18 at 17:11
  • $\begingroup$ That was just for ease of notation. The object is to prove that $Q$ is central in $N_N(Q)$ so that we can use Burnside's Transfer Theorem. So just write $N_N(Q) = QH$ where $H$ is a $p$-complement of $Q$ in $N$. I don't think I ever did assume that $G=PN$, but it is clear that you could, because we are trying to prove a result about $N$. $\endgroup$ – Derek Holt Mar 31 '18 at 18:08
  • $\begingroup$ @DerekHolt, Thank you for answer. But in this case the delicate moment will be with $H^g$ is another complement for $Q$ because $H^g$ is not necessarily subgroup of $N_N(Q)$. In the case $N=N_N(Q)$ there is no such a moment. What do you think? $\endgroup$ – Mikhail Goltvanitsa Mar 31 '18 at 18:35

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