Image under Epimorphism of Center is Subset of Center

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Theorem

Let $G$ and $H$ be groups.

Let $\theta: G \to H$ be an epimorphism.

Let $\map Z G$ denote the center of $G$.


Then:

$\theta \sqbrk {\map Z G} \subseteq \map Z H$


Proof

Let $y \in \theta \sqbrk {\map Z G}$.

Let $t \in H$.

We have that:

$y = \map \theta z$

for some $z \in \map Z G$

As $\theta$ is an epimorphism, it is by definition surjective.

Then:

$t = \map \theta s$

for some $s \in G$.

Hence:

\(\ds y t\) \(=\) \(\ds \map \theta z \map \theta s\)
\(\ds \) \(=\) \(\ds \map \theta {z s}\) Definition of Group Epimorphism
\(\ds \) \(=\) \(\ds \map \theta {s z}\) Definition of Center of Group
\(\ds \) \(=\) \(\ds \map \theta s \map \theta z\) Definition of Group Epimorphism
\(\ds \) \(=\) \(\ds t y\) Definition of Group Epimorphism

As $t$ is arbitrary, it follows that $y \in \map Z H$.

The result follows.

$\blacksquare$


Sources

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