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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 $N\cap Z(G)\neq \{1\}$ ?

Nicky Hekster
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Zihao huang
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  • To prove that the socle of $G$ is contained in the center, it suffices to prove that any minimal normal subgroup of $G$ is contained in the center. Do you see what happens if you assume the contrary? – Haran Jan 22 '23 at 04:38
  • If you are worried about the part of proving that $N \cap Z(G)$ is non-trivial, this is true for all non-trivial $p$-groups $G$ and non-trivial normal subgroups $N$ (this is where you have to use the fact that $G$ is a $p$-group). I would assume that this result must have appeared earlier in your book, but if not, it's a great exercise. Try using the class equation. – Haran Jan 22 '23 at 04:42
  • This follows trivially from the fact that non-trivial normal subgroups of $p$-groups intersect the center non-trivially, making this a repeat. I am tempted to use my golden hammer and close it as a duplicate. – Arturo Magidin Jan 22 '23 at 22:09

1 Answers1

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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 $soc(G) \subseteq Z(G)$.

Nicky Hekster
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