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Let $G$ be finite nilpotent. I want to show that each minimal normal subgroup of $G$ is contained in $Z(G)$ and has order $p$.

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We have that $Z(G)=\{g\in G\mid ga=ag , \ \forall a\in G\}$.

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$G$ is nilpotent iff $\exists$ a series of normal subgrousp $N_i$: $$1\leq N_1\leq N_2 \leq \dots \leq N_k=G$$ such that $N_{i+1}/N_i\subseteq Z(G/N_i)$, right?

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Could you give me some hints how we could show that each minimal normal subgroup of $G$ is contained in $Z(G)$ ?

Mary Star
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2 Answers2

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Suppose that $N$ is a minimal normal subgroup of $G$.

1.) Since $N$ is nontrivial and normal, $N ∩ Z(G)$ is nontrivial by nilpotency of $G$. Since $N ∩ Z(G)$ is a normal subgroup of $G$, by minimality of $N$ we have $N = N ∩ Z(G)$, so $N\subseteq Z(G)$.

2.) Since $N$ is in the center of $G$, every subgroup of $N$ is a normal subgroup of $G$, so $N$ contains no proper nontrivial subgroups. It follows that $N$ is abelian and simple, so it has prime order $p$.

Dietrich Burde
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  • We have that $G$ is nilpotent so $\exists N_i$, $N_i\trianglelefteq G$ with $N_{i+1}/N_i\subseteq Z(G/N_i)$, right? Why do we have that $N\cap Z(G)$ is nontrivial? – Mary Star Mar 27 '16 at 01:04
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    By the nilpotency of $G$. This is a result on nilpotent groups- see this MSE-question and this one. In particular, nilpotent groups have non-trivial center. – Dietrich Burde Mar 27 '16 at 09:17
  • What exactly do we mean by minimal normal subgroup? A group $G$ is called nilpotent iff $\exists$ a series of normal subgroups $$1\leq N_1\leq N_2\leq \dots \leq N_k=G$$ such that $N_{i+1}/N_i\subseteq Z(G/N_i)$, right? Is the minimal subgroup in that case $N_1$ ? – Mary Star Mar 28 '16 at 10:42
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    No, not necessarily, see here. – Dietrich Burde Mar 28 '16 at 12:24
  • Is every abelian and simple group, a $p$ group? – Mary Star Mar 29 '16 at 09:42
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    Yes, it is isomorphic to the cyclic group $C_p$ of prime order, see here. – Dietrich Burde Mar 29 '16 at 09:45
  • Ah ok... Do we show that $N$ is abelian as follows? $$$$ We have that $N\subseteq Z(G)={g\in G\mid ga=ag, \forall a\in G}$. So, for all $x,y\in N$ we have that $x,y\in Z(G)$ and $x,y\in G$. Therefore, $xy=yx$. That means that $N$ is abelian. $$$$ Could I improve something? – Mary Star Mar 29 '16 at 22:47
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Since $\;G\;$ is nilpotent, any non-trivial normal subgroup intersects non-trivially the center, and then for minimal non-trivial normal $\;N\lhd G\;$ we get

$$1<|N\cap Z(G)|\stackrel{\text{minim.}}\implies N\cap Z(G)=N\iff N\le Z(G)$$

DonAntonio
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  • We have that $G$ is nilpotent so $\exists N_i$, $N_i\trianglelefteq G$ with $N_{i+1}/N_i\subseteq Z(G/N_i)$, right? Why do we have that any non-trivial subgroup intersects non-trivially the center? – Mary Star Mar 27 '16 at 01:04
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    I really don't understand what you mean with your first question and what it has to do with your original one. As for the main claim in the answer: if we take the upper central series $;1=Z_0\le Z_1\le\ldots\le Z_n=G;$, then there is the smallest $;k;$ s.t. $;N\cap Z_k=1,,,,N\cap Z_{k+1}\neq1;$ , so take $;1\neq n\in N\cap Z_{k+1};$ , then for all $;x\in G;$: $$[n,x]\in[N\cap Z_{k+1},,G]\le [N\cap Z_k,,G]=1\implies n\in Z(G)$$ since $;[N,G]\le N;$ (normality), and also $;[Z_{k+1},G]\le Z_k;$ (Central series). – DonAntonio Mar 27 '16 at 05:20
  • I got stuck right now... What exactly do we mean by minimal normal subgroup? A group $G$ is called nilpotent iff $\exists$ a series of normal subgroups $$1\leq N_1\leq N_2\leq \dots \leq N_k=G$$ such that $N_{i+1}/N_i\subseteq Z(G/N_i)$, right? Is the minimal subgroup in that case $N_1$ ? – Mary Star Mar 28 '16 at 11:59
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    A normal subgroup $;N\lhd G;$ is minimal if it is not trivial ( i.e., $;N\neq{1};$) and there is no normal subgroup $;M;$ of $;G;$ such that $;N\subsetneq M\subsetneq G;$. Your definition of nilpotent is correct but a little odd, since it is usually defined by means of the upper central series, but it doesn't matter: the definition isn't connected with minimal/maximal subgroups, either normal or whatever. – DonAntonio Mar 28 '16 at 12:06
  • Ah ok... Is the intersection $N\cap Z(G)$ a normal subgroup of $G$ ? – Mary Star Mar 28 '16 at 13:19
  • @MaryStar Of course. The intersection of any two normal subgroups is a normal subgroup. – DonAntonio Mar 28 '16 at 16:20