Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Perhaps this theorem can shed some insight into why the smallest $p$ is useful?… – Ian Coley Apr 18 '13 at 17:49
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 $G$. – Tobias Kildetoft Apr 18 '13 at 17:53

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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