Prime power order implies center is normality-large

Symbolic statement
Let $$G$$ be a group of prime power order, and $$N$$ be a normal subgroup. Denote by $$Z(G)$$ the center of $$G$$.Then:

$$N \cap Z(G) = 1 \iff N = 1$$

Verbal statement
The center of a group of prime power order is a fact about::normality-large subgroup: its intersection with any nontrivial normal subgroup is nontrivial.

Related facts

 * prime power order implies not centerless: The center of a nontrivial group of prime power order is nontrivial.
 * any group of prime power order is nilpotent: Any group of prime power order is nilpotent.
 * Nilpotent implies center is normality-large: In a nilpotent group, the intersection of the center with any nontrivial normal subgroup is nontrivial.

Facts used

 * 1) uses::Fundamental theorem of group actions
 * 2) uses::Class equation of a group action
 * 3) uses::Lagrange's theorem

Proof
Given: A group $$G$$ of prime power order with center $$Z(G)$$, a nontrivial normal subgroup $$N$$ of $$G$$.

To prove: $$N \cap Z(G)$$ is nontrivial.

Proof: Consider the action of $$G$$ on $$N$$ by conjugation. This action is well-defined because $$N$$ is normal in $$G$$.

The fixed points under this action are the elements of $$N$$ that commute with every element of $$G$$ which is $$N \cap Z(G)$$.

Suppose, under this action, the elements of $$N$$ that are not fixed points decompose into orbits $$\mathcal{O}_1, \mathcal{O}_2, \dots, \mathcal{O}_r$$. Then, we have:

$$|N| = |N \cap Z(G)| + \sum_{i=1}^r |\mathcal{O}_i|$$.

By the fundamental theorem of group actions (fact (1)), we have:

$$|N| = |N \cap Z(G)| + \sum_{i=1}^r [G:G_i]$$,

where $$G_i$$ is the stablizer of some point in $$\mathcal{O}_i$$.

Since each $$\mathcal{O}_i$$ has size greater than one, $$G_i$$ is a proper subgroup, so we have that $$[G:G_i] = |G|/|G_i|$$ (by fact (3), Lagrange's theorem). Since the order of $$G$$ is a power of $$p$$ and $$G_i$$ is proper, $$|G/G_i|$$ is a power, and in particular, a multiple of $$p$$. Thus, we get:

$$|N| \equiv |N \cap Z(G)| \mod p$$

Since $$N$$ is nontrivial, $$|N|$$ is a multiple of $$p$$, hence so is $$|N \cap Z(G)|$$. Thus, $$N \cap Z(G)$$ is nontrivial.