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This question is from Hungerford Algebra Chapter Structure of groups.

If $G$ is a finite nilpotent group, then every minimal normal subgroup of $G$ is contained in $C(G)$ and has prime order.

A minimal normal subgroup of a group $G$ is a nontrivial normal subgroup that contains no proper subgroup which is normal in $G$, where $C(G )$ is normal subgroup of $G$.

$G$ is nilpotent means there exists $n$ such that $C_n (G) =\langle e\rangle$, $C_n(G)$ is the inverse image of $C(G/ C_{n-1}(G))$.

Let $H$ be a minimal normal subgroup. I am at loss of ideas and not even able to start and would appreciate hints for it.

Also, I am unable to form some reasons on why minimal subgroup must have prime order?

Can you please help with that?

Any help would be much appreciated.

Arturo Magidin
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  • Your definition of nilpotent is precisely backwards. Nilpotency means that there is an $n$ such that $C_n(G)=G$, not that it is trivial. – Arturo Magidin Sep 02 '21 at 20:37

1 Answers1

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  1. Any normal subgroup in the nilpotent group intersect the center non-trivially
  2. the intersection of two subgroup is still a subgroup. In particular, the intersection of a subgroup with the center is still normal.
SmoothKen
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  • Your phrasing of 2 is poor. The intersection of a subgroup with the center is normal because every subgroup of the center is normal, not just because it is a subgroup. – Arturo Magidin Sep 02 '21 at 20:38