A normal subgroup $N$ of $G$ with $\operatorname{gcd}(|N|,|G/N|)=1$

Question

Let $G$ be a finite group and $N$ be a normal subgroup of $G$ such that the centrilizer of $x$ in $G$ is a subset of $N$ for each $x \in N \setminus \{e\}$ ($\operatorname{C}_{G}(x) \subseteq N$, $\forall x \in N \setminus \{e\}$). Prove $\operatorname{gcd}(|N|,|G/N|)=1$.

Answer 1

Proceed by contradiction. Let $p$ be a prime that divides the order of both $N$ and $G/N$.

Let $P$ be a $p$-Sylow subgroup of $G$. Since $N$ is normal, $P \cap N$ is a Sylow $p$-subgroup of $N$, hence non-trivial. Moreover $P \cap N$ is a normal subgroup of $P$, and thus intersects non-trivially the centre $Z(P)$ of $P$.

So let $1 \ne x \in Z(P) \cap N$. The centralizer of $x$ contains the whole of $P$, which is not contained in $N$, as $p$ divides the order of $G/N$.