Theorem: Let G be a finite group of order n and let p be a prime. Write $n=p^am$ with $a \in\mathbb{N}^*$ and $m \in\mathbb{N}$ such that p does not divide m.
(a) If P is a subgroup of G of order $p^b$ with $0 \leq b <a$ then there exists a subgroup $P^{*}$ of G of order $p^{b+1}$ such that P is a normal subgroup of $P^{*}$
(b) For every $b \in {0,.....,a}$ there exists a subgroup P of G with $|P|=p^b$
Proof (a) Assume that P is a subgroup of G of order $p^b$ with $1 \leq b<a$. Then p divides $[G:P]$ and, by Corollary 12.3, p also divides $[N_G(P):P]=|N_G(P)/P|$. By Cauchy's Theorem, the group $N_G(P)/P$ has a subgroup of order p. By the Correspondence Theorem this subgroup must be of the form $P^{*}/P$ with $P \leq P^{*} \leq N_G(P).$ This implies that P is a normal subgroup of $P^{*}$ and $|p^{*}|=|P|*|P^{*}/P|=p^b*p=p^{b+1}$
b) This follows immediately from part (a) by induction on b
Corollary 12.3: Let G be a finite group and let $P$ be a p-subgroup of G such that p divides $[G:P]$. Then p divides $[N_G(P):P]$
Ok, so in the first sentence. How do we infer p divides $[G:P]$. Can you explain in detail the theorems that were used to prove this and how they relate to the proof?.
I think now I understand the proof up to the correspondence theorem
Thank you
Note that $[G:P]=\frac{|G|}{|P|}=\frac{p^am}{p^b}=p^{(a-b)}m$. Since $a-b>0$, then $p|p^{(a-b)}|p^{(a-b)}m=[G:P].$ So $p|[G:P]$.
Consider the canonique projection $\pi:N_G(P)\longrightarrow\frac{N_G(P)}{P}$. Applying the correspondence theorem we see that every subgroup of $Im(\pi)=\frac{N_G(P)}{P}$ has the form $\frac{P^*}{P}$ where $P^*$ is a subgroup of $N_G(P)$.