2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @Geoff: ypu're right, I've taken my comment down. $\endgroup$ – Chris Godsil Jan 14 '13 at 13:09
2
$\begingroup$

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 $N_G(H)\setminus H$.

$\endgroup$
4
$\begingroup$

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 $2$.

$\endgroup$
  • $\begingroup$ good work, like usual! +1 $\endgroup$ – Namaste Feb 18 '13 at 0:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.