# Center of normalizer of cyclic group

Given a Group of order $pm$ with $p$ the least prime dividing the order of $G$ and p does not divide $m$. Is it true that a $p$-sylow subgroup is contained in the center of its normalizer.

I think yes, because of the Action by conjugation of normalizer on a $p$ sylow, and from the property of p. But i'm not sure of this.

Edit: I write here my proof of this fact

The notation I'll use is the following $Z_p$ for the p-Sylow subgroup (we know it's cyclic for the square free hypothesis), $N_p$ the normalizer of the p-Sylow and $Z(N_p)$ for his center. $|.|$ stands for the cardinality of $"."$.

So we have that $Z_p$ is normal in $N_p$ then we have a "natural" action of $N_p$ on $Z_p$ via conjugation, so we have that, given $x \in Z_p$ its orbit $Orb(x)$ will have a cardinality $|Orb(x)||Stab(x)|=|N_p|$ . But the cardinality of $Orb(x)$ can be only 1 or p because of the minimality of p (and because $Z_p$ is a subgroup of $N_p$) then if there exists one orbit with cardinality p then we know,this holds in general, that the $Stab(x)$ will contain $Z_p$ because it is abelian, and then it will have cardinality $|Stab(x)|=p\times other things"$ And then we will have that $|N_p|$ is not square free and so this is an absurd.

Let $P \in Syl_p(G)$. Note that $P \cong C_p$ and that $N_G(P)/C_G(P)$ injects homomorphically into $Aut(P) \cong C_{p-1}$. Since $p$ is the smallest prime dividing the order of $G$, we must have $N_G(P)=C_G(P)$. It now follows that $P \subseteq Z(N_G(P))$.