$H$ of order $p$ normal in $G$ , g.c.d.$(|G|,p-1)=1$ , to prove that $H \subseteq Z(G)$

Question

If $G$ is a finite group and $H$ is a normal subgroup of $G$ of order $p$(prime) such that

g.c.d.$(|G|,p-1)=1$ , then how to prove that $H \subseteq Z(G)$ ?

Please don't use any Sylow theorem or Cauchy theorem , I want solution with Homorphism techniques (extended Cayley's theorem is also allowed) . Thanks in advance

Answer 1

Ok, I'll have a try following the hint of @James. Since $|H|=p$ with $p$ prime, we have $H \simeq \mathbb{Z}/p\mathbb{Z}$ and $\mathrm{Aut}(H) \simeq \mathbb{Z}/(p-1)\mathbb{Z}$. Let $n$ be the order of $G$.

Following the hint, we have that $G/C_{G}(H)$ embeds in $\mathrm{Aut}(H)$, and so $[G:C_{G}(H)] | p-1$ ; but it is also known that $[G:C_{G}(H)] | n $, therefore $[G:C_{G}(H)]=1$, which is the statement we're aiming at.

Am I doing it right ?

edit : proof of the hint. If $H$ is normal in $G$, then for every $g$ we have $ghg^{-1} \in H$, so we can define $\phi : G \to \mathrm{Aut}(H)$ by $$\phi : g \to \big(h \to ghg^{-1} \big)$$.

This a morphism which kernel is precisely $C_{G}(H)$, so it factorizes in a morphism $\tilde{\phi} : G/C_{G}(H) \to \mathrm{Aut}(H)$.

Answer 2

Hint: If $C_{G}(H)$ denotes the centraliser of $H$ in $G$, then $G/C_{G}(H)$ embeds in $\operatorname{Aut}(H)$.