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

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Have you considered what $N\cap Z(G)$ might look like if $N$ were a minimal normal subgroup? – user641 Nov 8 '10 at 16:45
First, the minimal normal subgroup is a abelian p-group, consider the intersection of the minimal normal subgroup and the commutator subgroup of G is the identity. – Yuan Nov 8 '10 at 16:51

1 Answer 1

up vote 3 down vote accepted

A finite nilpotent group is a product of $p$-groups. So you can do a very quick computation to show that you can reduce to the case where $G$ is a $p$-group.

Then look at this question to answer Steve D's comment query.

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Well, we can get by with more generality by considering that eventually [G,G,...,N] must be central, and contained in N. This shows the question is true even if we remove the word "finite". – user641 Nov 8 '10 at 17:25
@Steve D: Agreed. – Arturo Magidin Nov 8 '10 at 17:26

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