4
$\begingroup$

Let $p$ be the smallest prime dividing the order of a finite group $G$. If $P$ in $\operatorname{Syl}_p(G)$ and $P$ is cyclic, prove that $N_G(P)=C_G(P)$.

This is not homework. It is from Dummit and Foote. I'm not sure how to apply that $p$ has the smallest order.

$\endgroup$
  • $\begingroup$ Perhaps this theorem can shed some insight into why the smallest $p$ is useful? groupprops.subwiki.org/wiki/… $\endgroup$ – Ian Coley Apr 18 '13 at 17:49
  • 5
    $\begingroup$ Hint: Use that $N_G(P)/C_G(P)$ is isomorphic to a subgroup of $\rm{Aut}(P)$ which has order $p-1$ which by assumption is coprime to the order of $G$. $\endgroup$ – Tobias Kildetoft Apr 18 '13 at 17:53
4
$\begingroup$

Take $\Sigma:N_G(P)\rightarrow \operatorname{Aut}(P)$ by $\Sigma(g)=\sigma_g:x\mapsto g^{-1}xg$. Then $N_G(P)/C_G(P)$ is isomorphic to $\Sigma[N_G(P)]$. Since $P$ is cyclic $\operatorname{Aut}(P)$ has order $\varphi(p^n)=p^{n-1}(p-1)$ where $p^n=|P|$. Furthermore, $P$ centralizes itself, so $\Sigma[P]=1$. All other subgroups of $N_G(P)$ must have order that does not divide $p^{n-1}(p-1)$, as by assumption all other primes are greater than $p$. Thus $\Sigma[N_G(P)]=1$. This completes the proof.

$\endgroup$

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.