Normal Subgroup of p-Group of Order p is Subset of Center

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 $21$