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$G$ is a finite group and $N$ is a normal subgroup of $G$. Let $H/N$ be any nontrivial subgroup of $G/N$ of prime power order. Then we have $|H/N|=p^n$, for some prime $p$ and $n\geq 1$. Let $P$ denote a Sylow $p$-subgroup of $H$. Applying the Frattini argument, we have ${\color{red}{N_{G/N}(H/N)=N_G(P)N/N.}}$

I’m reading a paper and I saw the statement above. I tried to prove it, but I don’t know if my proof is right. The following is my attempt in detail:

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$\overline{G}:=G/N$. Now I want to prove $N_{\overline G}(\overline H)= \overline{N_{G}(P)}$.

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(i) ${\color{red}{N_{\overline G}(\overline P)= \overline {N_G(P)}}}$

By the definition of normalizers, $$\begin{array}{rcl} \overline {N_G(NP)} &=& \{\overline g \in \overline G \mid gNPg^{-1} = NP \} \\ &=& \{\overline g \in \overline G \mid \overline {gNPg^{-1}} = \overline {NP} \} \\ &=& \{\overline g \in \overline G \mid \overline {gPg^{-1}} = \overline P \} \\ &=& N_{\overline G}(\overline P). \end{array}$$

$H/N$ is well-defined, so $N$ is contained in $H$. Therefore $NP$, which contains $P$, is a subgroup of $H$. Since $P$ is a Sylow $p$-subgroup of $H$, $P$ is also a Sylow $p$-subgroup of $NP$. By the Frattini argument, $$N_G(NP)=NP~N_{N_G(NP)}(P).$$

For all $x\in N_G(P)$, we have $$x^{-1}(NP)x=(x^{-1}N)Px=(Nx^{-1})Px=N(x^{-1}Px)=NP,$$ which gives $x\in N_G(NP)$. The inclusion $N_G(P)\subseteq N_G(NP)$ actually implies that no element out of $N_G(NP)$ in $G$ can normalize $P$. In other words, searching $G$ for an element normalizing $P$ is equivalent to searching $N_G(NP)$ for an element normalizing $P$, i.e.,$$N_G(P)=N_{N_G(NP)}(P).$$

Therefore, $$N_G(NP)=NP~N_{N_G(NP)}(P)=NP~N_G(P)=N~N_G(P).$$

Thus, $$N_{\overline G}(\overline P)=\overline {N_G(NP)}=\overline {N~N_G(P)}=\overline {N_G(P)}.$$

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(ii) ${\color{red}{\overline{P}=\overline{H}}}$

Since $P$ is a Sylow $p$-subgroup of $H$, $\overline{P}$ is a Sylow $p$-subgroup of $\overline{H}$. I show that in the following way. From $PN/N\cong P/(P\cap N)$, we know \begin{align} |PN/N|&=|P/(P\cap N)|\\ |PN/N|•|P\cap N|&=|P|. \end{align} Then$$\dfrac{|H|}{|P|}=\dfrac{|H|}{\frac{|PN|}{|N|}•|P\cap N|}=\dfrac{|H|}{|PN|}•\dfrac{|N|}{|P\cap N|}=[H:PN]•[N:P\cap N].$$ Since $P$ is a Sylow $p$-subgroup of $H$, $\frac{|H|}{|P|}$ is not divisible by $p$. Hence the divisor $\frac{|H|}{|PN|}$ of $\frac{|H|}{|P|}$ is not divisible by $p$. It implies that $[H/N:PN/N]=\frac{|H|}{|PN|}$ is not divisible by $p$.

$PN/N\cong P/(P\cap N)$ is a $p$-subgroup of $H/N$ and now we know that $|PN/N|$ is the largest $p$-power dividing $|H/N|$. Therefore $PN/N$ is a Sylow $p$-subgroup of $H/N$.

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$\overline{H}$ is a $p$-group, which means $\overline{H}$ is the only Sylow $p$-subgroup of itself. It gives $\overline P=\overline H$ and thus $N_{\overline G}(\overline P)= N_{\overline G}(\overline H)$.

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Hence we conclude that $N_{\overline G}(\overline H)= \overline{N_{G}(P)}$.

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PS: I’m worried about the last part most, where I claimed $\overline{P}=\overline{H}$. It just seems a little bit unbelievable to me... I struggled with that point for hours, but I suddenly found it’s so clear. Is it really that easy? Thanks!

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  • $\begingroup$ Your argument is correct. For your last part, in general, if $P$ is a Sylow $p$-subgroup of $G$ and $N$ is normal subgroup, then $PN/N$ is a Sylow $p$-subgroup of $G/N$. $\endgroup$ – Nicky Hekster Jan 17 at 19:07

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