Is it true to say $Z(G)\subseteq N_G(H)$?

Question

Let $p$ be an prime number and $G$ a group of order $p^n$ and $H$ be subgroup of $G$ of order $p^{n-2}$ and which is not normal in $G$. Is it true to say that $Z(G)\subseteq N_G(H)$?

Answer 1

Let's take it step by step. If $G$ is any group and $H$ is any subgroup of $G$, then

1. by definition of $\mathcal{Z}(G)$, $zg=gz$ for any $z\in \mathcal{Z}(G)$ and $g\in G$.

2. Thus, for any $h\in H$, $zh=hz$,

3. so $z^{-1}hz=h$,

4. so $z^{-1}Hz=H$,

5. so $\mathcal{Z}(G)\leqslant \mathcal{N}_G(H)$.

6. In fact, we can say even that $\mathcal{Z}(G)\leqslant \mathcal{C}_G(H)$, which is itself a subgroup of $\mathcal{N}_G(H)$ (why?).

Answer 2

The answer by Alexander Gruber is clear; I will add a comments: we can make a more general statement than $Z(G)\subseteq N_G(H)$.

The centralizer of $H$ is $G$ is the set $C_G(H)=\{g\in G\colon gh=hg \forall h\in H \}$, the elements of $G$ commuting pointwise with elements of $H$. Clearly, $C_G(H)\leq N_G(H)$ (in fact $C_G(H)\triangleleft N_G(H)$). And then, some part of proof of Alexander says that $Z(G)\leq C_G(H)$.