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:
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 $j \in \set {1, 2, \ldots, m}$.
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Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $21$