Central Subgroup is Normal

Theorem

Let $G$ be a group.

Let $H$ be a central subgroup of $G$.

Then $H$ is a normal subgroup of $G$.

Proof 1

Let $H$ be a central subgroup of $G$.

By definition of central subgroup:

$H \subseteq \map Z G$

where $\map Z G$ is the center of $G$.

Thus we have that $H$ is a group which is a subset of $\map Z G$.

Therefore by definition $H$ is a subgroup of $\map Z G$.

We also have from Center of Group is Abelian Subgroup that $\map Z G$ is an abelian group.

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It follows from Subgroup of Abelian Group is Normal that $Z$ is a normal subgroup of $G$.

$\blacksquare$

Proof 2

Let $H$ be a central subgroup of $G$.

By definition of central subgroup:

$H \subseteq \map Z G$

where $\map Z G$ is the center of $G$.

Then:

\(\ds \forall x \in G: \forall h \in H: \, \)

\(\ds x h x^{-1}\)

\(=\)

\(\ds x x^{-1} h\)

as $h \in H \implies h \in \map Z G$

\(\ds \)

\(=\)

\(\ds h\)

\(\ds \leadsto \ \ \)

\(\ds x h x^{-1}\)

\(\in\)

\(\ds H\)

as $h \in H$

\(\ds \leadsto \ \ \)

\(\ds H\)

\(\lhd\)

\(\ds G\)

Definition of Normal Subgroup

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures