Center is a Subgroup

Theorem

The center $Z \left({G}\right)$ of any group $G$ is a subgroup of $G$.

Proof 1

For brevity, suppress the symbol for the group operation (which may be $\circ$, or $+$).

Apply the Two-step Subgroup Test:

Condition $(1)$

By the definition of identity, $e g = g e = g$ for all $g \in G$.

So, $e \in Z \left({G}\right)$, meaning $Z \left({G}\right)$ is nonempty.

$\Box$

Condition $(2)$

Suppose $a, b \in Z \left({G}\right)$.

Using the associative property and the definition of center, we have:

$\forall g \in G: \left({a b}\right) g = a \left({b g}\right) = a \left({g b}\right) = \left({a g}\right) b = \left({g a}\right) b = g \left({a b}\right)$

Thus, $a b \in Z \left({G}\right)$.

$\Box$

Condition $(3)$

Suppose $c \in Z \left({G}\right)$. Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle c\)	\(\in\)	\(\displaystyle Z \left({G}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \forall g \in G: c g\)	\(=\)	\(\displaystyle g c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of center
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle c^{-1} \left({c g}\right) c^{-1}\)	\(=\)	\(\displaystyle c^{-1} \left({g c}\right) c^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({c^{-1} c}\right) g c^{-1}\)	\(=\)	\(\displaystyle c^{-1} g \left({c c^{-1} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle e g c^{-1}\)	\(=\)	\(\displaystyle c^{-1} g e\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle g c^{-1}\)	\(=\)	\(\displaystyle c^{-1} g\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle c^{-1}\)	\(\in\)	\(\displaystyle Z \left({G}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\Box$

Therefore, $Z \left({G}\right) \le G$.

$\blacksquare$

Proof 2

We have the result Center is Intersection of Centralizers.

That is, $Z \left({G}\right)$ is the intersection of all the centralizers of $G$.

All of these are subgroups of $G$ by Centralizer of Group Element is Subgroup.

Thus from Intersection of Subgroups, $Z \left({G}\right)$ is also a subgroup of $G$.

$\blacksquare$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 35 \delta$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 37.2$