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, $e \in \map Z G$, meaning $\map Z G$ is non-empty.

Condition $(2)$
Suppose $a, b \in \map Z G$.

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


 * $\forall g \in G: \paren {a b} g = a \paren {b g} = a \paren {g b} = \paren {a g} b = \paren {g a} b = g \paren {a b}$

Thus, $a b \in \map Z G$.

Condition $(3)$
Suppose $c \in \map Z G$. Then:

Therefore:
 * $\map Z G \le G$