Center is Characteristic Subgroup

Theorem

Let $G$ be a group.

Then its center $\map Z G$ is characteristic in $G$.

Proof

By Identity Mapping is Group Automorphism, there exists at least one automorphism of $G$.

Let $\phi$ be an automorphism of $G$.

Let $x \in \map Z G, y \in G$.

Then:

\(\ds \map \phi x y\)

\(=\)

\(\ds \map \phi x \map \phi {\map {\phi^{-1} } y}\)

automorphisms are bijections

\(\ds \)

\(=\)

\(\ds \map \phi {x \map {\phi^{-1} } y}\)

Definition of Group Homomorphism

\(\ds \)

\(=\)

\(\ds \map \phi {\map {\phi^{-1} } y x}\)

Definition of Center of Group

\(\ds \)

\(=\)

\(\ds \map \phi {\map {\phi^{-1} } y} \map \phi x\)

Definition of Group Homomorphism

\(\ds \)

\(=\)

\(\ds y \map \phi x\)

Hence $\map \phi x \in \map Z G$.

So we have $\phi \sqbrk {\map Z G} \subseteq \map Z G$.

Since $\phi^{-1}$ is also an automorphism:

$\phi^{-1} \sqbrk {\map Z G} \subseteq \map Z G$

Since $\phi$ is a bijection:

$\map Z G = \phi \sqbrk {\phi^{-1} \sqbrk {\map Z G}} \subseteq \phi \sqbrk {\map Z G}$

Therefore we conclude that $\phi \sqbrk {\map Z G} = \map Z G$.

Hence $\map Z G$ is characteristic in $G$.

$\blacksquare$