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