If $G$ is a finite $p$-group, then the socle of $G$ is contained in the centre $Z(G)$ This is an exercise in Permutation Group by D.Dixon and Brian Mortimer, from page112, exercise 4.3.3.

Recall: The socle of $G$ is the subgroup generated by the set of all minimal normal subgroups of $G$.

I have known that if $G$ is a finite $p$-group, then $Z(G)$ is a nontrivial normal subgroup of $G$, and if $H$ is a normal subgroup of $G$ with order $p$, then $H$ must be contained in $Z(G)$ by the $N/C$ theorem. My question is, given an arbitrary minimal normal subgroup $N$ of $G$, how can I show that $N\cap Z(G)\neq \{1\}$ ?
 A: Step 1 Let $G$ be a $p$-group and $1 \lt N \unlhd G$. Then $N \cap Z(G) \gt 1$.

 Proof $G$ acts on $N$ by conjugation, which is well-defined since $N$ is normal. Now, obviously $\{1\}$ is a single orbit under this action and the other orbits have a length a power of $p$, since $G$ is a $p$-group. If the rest of the orbits would have a length larger than $1$, then - applying the orbit-stabilizer theorem - we reach a contradiction, since $|N| \equiv 0$ mod $p$ and $N=\{1\} \cup \mathscr{O}_2 \cup \cdots \cup \mathscr{O}_r$ and the cardinality of the right hand-side $\equiv 1$ mod $p$. Hence there must be another orbit of length $1$, that is an $n \in N-\{1\}$ fixed by $G$, so $n \in Z(G)$.$\square$


Step 2 Let $G$ be a $p$-group, then any minimal normal subgroup $N$ is contained in $Z(G)$. And hence $soc(G) \subseteq Z(G)$.

 Proof By Step 1, $N \cap Z(G) \gt 1$, but of course $N \cap Z(G) \unlhd N$. Since $N$ is minimal normal we must have $N \cap Z(G)=N$, equivalent to $N \subseteq Z(G)$. $\square$


The statement can be generalized as follows: let $G$ be a nilpotent group, then $soc(G) \subseteq Z(G)$.
