Center of Group is Subgroup

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

Direct Proof
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.

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

Condition $(3)$
Suppose $c \in Z \left({G}\right)$. Then:

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

Proof 2
The definition of centralizer implies that $x \in Z \left({G}\right)$ iff $x$ is in the centralizer of all elements of $G$.

Thus $Z \left({G}\right)$ 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 $G$.