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.

By definition:

$\map Z G = G$

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

This needs considerable tedious hard slog to complete it. In particular: Needs more thought. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $21$