Frattini's argument and normalizers Dummit chapter 6

Question

At chapter 6.1, there is exercise 20, D&F Abstract Algebra 3rd E,

which is the following:

Let $p$ be a prime and $P$ be a p-subgroup of the finite group $G$, let $N$ be a normal subgroup of $G$ whose order is relatively prime to $p$ and let $\bar{G}=G/N$. Prove the following:

(a) $N_\bar{G}(\bar{P})=\overline{N_G(P)}$.

(b) $C_\bar{G}(\bar{P})=\overline{C_G(P)}$.

After a bit of trying to solve this problem, I checked other's solution at mathexchange here.

But the problem posted looks different , cause it assumes that $P$ is a Sylow p-subgroup of G. and they were showing that $N_\bar{G}(\overline{PN})=\overline{N_G(P)N}$. (all the upper bar indicate that these are quotient subgroup of $G/N$.)

(1) I don't think they are actually same problem, because of the above assumption that $P$ is a Sylow p-subgroup of G and what they are trying to show. Is it right?

(2) And I couldn't figure out how to apply the Frattini's argument to this problem, could you give me a hint? and then I'd like to solve this on my own.

Answer 1

Since you are only asking for a hint, I will just point you in the right direction on (b):

(a) No, they aren't precisely the same problem, but they are strongly related as my hint for (b) will show.

(b) The proof for exercise 20 is going to involve looking at the group $PN$, and in that group $P$ is a Sylow subgroup. Elements from $G$ acting by conjugation either normalize $PN$ or they don't. For the ones that do, they take $P$ to another Sylow subgroup of $PN$ (and here you use the Frattini argument). For the ones that don't, then since $N$ is normal, they conjugate $P$ to a subgroup that is not contained in $PN$ (and here you use the correspondence theorem).

Answer 2

(a) Consider $gN\in N_{\overline{G}}\left(\overline{P}\right)$ so $g^{-1}N\overline{P}gN=\overline{P}$. That is, let $\overline{P}=\left\langle xN\right\rangle$ so $g^{-1}xgN=x^{k}N$. Then there is the sequence of inclusions \begin{align*} g &\in N_{G}\left(P\right)\\ gN &\in\overline{N_{G}\left(P\right)}\\ N_{\overline{G}}\left(\overline{P}\right) &\leq\overline{N_{G}\left(P\right)} \end{align*} On the other hand, $PN/N$ is a Sylow subgroup of $PN/N$ which is a normal subgroup of $\overline{N_{G}\left(P\right)}$ so Frattini's Argument implies \begin{align*} \overline{N_{G}\left(P\right)} &=\left(PN/N\right)N_{\overline{N_{G}\left(P\right)}}\left(\overline{P}\right) =N_{\overline{N_{G}\left(P\right)}}\left(\overline{P}\right) \leq N_{\overline{G}}\left(\overline{P}\right) \end{align*} It must be that $N_{\overline{G}}\left(\overline{P}\right)=\overline{N_{G}\left(P\right)}$. Thus the result.

(b) It is obvious that \begin{align*} \overline{N_{G}\left(P\right)}/\overline{C_{G}\left(P\right)} &\cong N_{G}\left(P\right)/C_{G}\left(P\right) \cong N_{\overline{G}}\left(\overline{P}\right)/C_{\overline{G}}\left(\overline{P}\right) \end{align*} and it must be that $C_{\overline{G}}\left(\overline{P}\right)=\overline{C_{G}\left(P\right)}$. Thus the result.