Center of Group is Subgroup/Proof 1

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