Nilpotent implies center is normality-large

This article gives the statement, and possibly proof, of the fact that in any nilpotent group, the subgroup obtained by applying a given subgroup-defining function (i.e., center) always satisfies a particular subgroup property (i.e., normality-large subgroup)
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This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., nilpotent group) must also satisfy the second group property (i.e., group whose center is normality-large)
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Statement

Verbal statement

In a nilpotent group, the center is a normality-large subgroup; in other words, the intersection of the center with any nontrivial normal subgroup is a nontrivial normal subgroup.

Related facts

Similar facts

• Prime power order implies center is normality-large

Generalizations

• Nilpotent implies intersection of normal subgroup with upper central series is strictly ascending till the subgroup is reached

Applications

• Nilpotent and non-abelian implies center is not complemented
• Minimal normal implies central in nilpotent
• Socle equals Omega-1 of center in nilpotent p-group
• Formula for number of minimal normal subgroups of group of prime power order
• Congruence condition relating number of normal subgroups containing minimal normal subgroups and number of normal subgroups in the whole group
• Thompson's critical subgroup theorem

Analogues in other algebraic structures

• Nilpotent implies center is ideal-large

Proof

Given: A nilpotent group $G$ with center $Z$. A nontrivial normal subgroup $N$ of $G$.

To prove: $N \cap Z$ is nontrivial.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Denote by $Z^{(i)}(G)$ the $i^{th}$ member of the upper central series of $G$. So $Z^{(0)}(G)$ is trivial, $Z^{(1)}(G) = Z$ is the center, and $Z^{(c)}(G) = G$ where $c$ is the nilpotency class of $G$.		$G$ is nilpotent.
2	There exists some $i \ge 0$ such that $N \cap Z^{(i)}(G)$ is trivial and $N \cap Z^{(i + 1)}(G)$ is nontrivial.		$N$ is nontrivial.	Step (1)	[SHOW MORE] We have $N \cap Z^{(0)}(G)$ is trivial and $N \cap Z^{(c)}(G) = N \cap G = N$ is nontrivial. Further, we know that since $Z^{(0)}(G) \le Z^{(1)}(G) \le \dots \le Z^{(c)}(G)$, we have $N \cap Z^{(0)}(G) \le N \cap Z^{(1)}(G) \le \dots N \cap Z^{(c)}(G)$. Thus, there is a last $i$ for which $N \cap Z^{(i)}(G)$ is trivial.
3	$[G,Z^{(i + 1)}(G)] \le Z^{(i)}(G)$.	Definition of upper central series
4	$[G,N] \le N$.		$N$ is normal in $G$.
5	$[G,N \cap Z^{i+1}(G)] \le [G,N] \cap [G,Z^{(i+1)}(G)$.			By definition of commutator of two subgroups in terms of the commutators being a generating set.
6	$[G,N \cap Z^{(i+1)}(G)]$ is trivial.			Steps (2),(3),(4),(5)	[SHOW MORE] Plugging Steps (3) and (4) into Step (5), we get $[G,N \cap Z^{i+1}(G)] \le N \cap Z^{(i)}(G)$. By the choice of $i$ from Step (2), $N \cap Z^{(i)}(G)$ is trivial, so $[G,N \cap Z^{i+1}(G)]$ is trivial.
7	$i = 1$, and $N \cap Z^{(1)}(G) = N \cap Z$ is nontrivial.			Steps (2), (6)	<toggledisplay>By Step (6), $N \cap Z^{(i+1)}(G)$ is contained in $Z = Z^{(1)}(G)$, so we have a nontrivial subgroup contained in both $N$ and $Z$. So, $N \cap Z$ is nontrivial. Reviewing the definition of $i$ from Step (2), we obtain that $i$ must equal 1.