Intersection of Normal Subgroup with Center in p-Group

Theorem
Let $p$ be a prime number

Let $G$ be a $p$-group.

Let $N$ be a non-trivial normal subgroup of $G$.

Let $\map Z G$ denote the center of $G$.

Then:
 * $N \cap \map Z G$ is a non-trivial normal subgroup of $G$.

Proof
First we note that:
 * Center of Group is Normal Subgroup

and from Intersection of Normal Subgroups is Normal:
 * $N \cap \map Z G$ is normal in $G$.

Suppose $G$ is abelian.

From Group equals Center iff Abelian:
 * $\map Z G = G$

Then:
 * $N \cap \map Z G = N$

which is non-trivial.

From Prime Group is Cyclic and Cyclic Group is Abelian, this will always be the case for $r = 1$.

So, suppose $G$ is non-abelian.

From Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is non-trivial.

From Union of Conjugacy Classes is Normal:


 * $N = \ds \bigcup_{x \mathop \in N} \conjclass x$

Let $\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$ be the conjugacy classes into which $N$ is partitioned.

From Conjugacy Class of Element of Center is Singleton, all of these will have more than one element.

From Number of Conjugates is Number of Cosets of Centralizer:
 * $\order {\conjclass {x_j} } \divides \order G$

for all $j \in \set {1, 2, \ldots, m}$.