$p$-group and normalizer

Question

Here is the question:

a) Show that if $p$ is a prime number and $P$ is a $p$-subgroup of a finite group $G$, then $[G:P]=[N_G(P):P]$(mod p), where $N_G(P)$ denotes the normalizer of $P$ in $G$.

b) Assume that $H$ is a subgroup of a finite group $G$, and that $P$ is a Sylow $p$-group of $H$. Show that if $N_G(P) \subset H$ then $P$ is a Sylow p-subgroup of $G$.

I have forgotten too much group theory to make much progress on this. I try to look at the action of $P$ on the set of cosets, $G/P$, but I must be missing something because not much connects to anything.

Any help would be very much appreciated.

Answer 1

Hint. Show that $p$ divides $[G:N_G(P)]$ and use $[G:P]=[G:N_G(P)][N_G(P):P]$.

If $P$ is a Sylow $p$-subgroup, then we are done. If not, then $P$ is not self-normalizing, and in fact is normalized by some subgroup of order $p|P|$ by the first Sylow theorem.

It should be noted that this is how computer algebra systems find Sylow subgroups, by taking elements of $p$-power order in $N_G(H)\setminus H$.

Answer 2

You can use the result of $1$ to see that $P$ is a $p$-sylow subgroup of finite group $G$. Let $|G|=p^{\alpha}m,(p,m)=1$.

• $H\leq G$ so $|H|=p^{s}n,~ s\leq\alpha,~~n\mid m$.

• $P$ is a sylow-$p$ subgroup of $H$ so $|P|=p^s$.

• $N_G(P)\subseteq H$ so $|N_G(P)|=p^sk,~~k\mid n$.

For all above $m,~n$ and $k$ are free of $p$ that is $(p,m)=1,~(p,n)=1,~(p,k)=1$.

Now apply the congruent relation resulted by $1$ for what we have:

$$[G:P]\equiv [N_G(P):P]~~~(\text{mod}~~ p)$$ or since $G$ is finite: $$|G|/|P|\equiv |N_G(P)|/|P|~~~(\text{mod}~~ p)$$ or $$p^{\alpha}m/p^s\equiv p^s.k/p^s~~~(\text{mod}~~ p)$$ or $$p^{\alpha-s}m\equiv k~~~(\text{mod}~~ p)$$

The latter makes a contradiction unless $\alpha=s$ and $m=k$. This is what you wanted in $2$.

Answer 3

Used lemma: For any subgroup $H \subset P$, $P$ is a p-group, $N(H)>H$

This lemma can be found in Dummit and Foote Chapter 6 Section 3.

(a) Now if $P$ is Sylow p-group, by third Sylow's theorem, $G:N_G(P) = 1$ mod p. $$G:P=G:N_G(P) \cdot N_G(P):P$$ we are done.

Otherwise, take $H$ such that $P \subset H$, H is Sylow p-subgroup of G. Then obviously $G:P=0$ mod p

Now, $N_G(P):P = N_G(P):N_H(P) \cdot N_H(P):P$, but by the lemma $p |N_H(P):P$, hence we are done.

(b) Assume for contradiction that P is not a Sylow p-subgroup of G, then there exists a Sylow p-subgroup $P^*$ such that $P<N_{P^*}(P) \subset P^*$.

However, note that $N_{P^*}(P) \subset N_G(P) \subset H$

Hence $N_{P^*}(P)$ is a p-group contained in H and strictly larger than P, contradiction since P is a Sylow p-subgroup of H.