Center of Group is Normal Subgroup/Proof 2

Theorem

Let $G$ be a group

The center $\map Z G$ of $G$ is a normal subgroup of $G$.

Proof

We have:

$\forall a \in G: x \in \map Z G^a \iff a x a^{-1} = x a a^{-1} = x \in \map Z G$

Therefore:

$\forall a \in G: \map Z G^a = \map Z G$

and $\map Z G$ is a normal subgroup of $G$.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 50$