A question about the Frattini argument and normalizers

$$G$$ 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!

• 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 $G/N$. – Nicky Hekster Jan 17 at 19:07