Normal Subgroup of p-Group of Order p is Subset of Center
Jump to navigation
Jump to search
Theorem
Let $p$ be a prime number.
Let $G$ be a $p$-group.
Let $N$ be a normal subgroup of $G$ of order $p$.
Then:
- $N \subseteq \map Z G$
where $\map Z G$ denotes the center of $G$.
Proof
From Intersection of Normal Subgroup with Center in p-Group:
- $\order {N \cap \map Z G} > 1$
From Intersection of Subgroups is Subgroup, $N \cap \map Z G$ is a subgroup of $N$.
It follows from Lagrange's Theorem that:
- $\order {N \cap \map Z G} = p$
and so:
- $N \cap \map Z G = N$
But from Intersection of Subgroups is Subgroup, $N \cap \map Z G$ is a subgroup of $\map Z G$
That is:
- $N$ is a subgroup of $\map Z G$
and the result follows.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $21$