# Center of Group is Subgroup

## Theorem

Let $G$ be a group.

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

$\Box$

### Condition $(2)$

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

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

 $\ds \forall g \in G: \,$ $\ds \paren {a b} g$ $=$ $\ds a \paren {b g}$ Definition of Associative Operation $\ds$ $=$ $\ds a \paren {g b}$ Definition of Center of Group $\ds$ $=$ $\ds \paren {a g} b$ Definition of Associative Operation $\ds$ $=$ $\ds \paren {g a} b$ Definition of Center of Group $\ds$ $=$ $\ds g \paren {a b}$ Definition of Associative Operation

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

$\Box$

### Condition $(3)$

Suppose $c \in \map Z G$. Then:

 $\ds c$ $\in$ $\ds \map Z G$ $\ds \leadsto \ \$ $\ds \forall g \in G: \,$ $\ds c g$ $=$ $\ds g c$ Definition of Center of Group $\ds \leadsto \ \$ $\ds c^{-1} \paren {c g} c^{-1}$ $=$ $\ds c^{-1} \paren {g c} c^{-1}$ Group Axiom $\text G 0$: Closure $\ds \leadsto \ \$ $\ds \paren {c^{-1} c} g c^{-1}$ $=$ $\ds c^{-1} g \paren {c c^{-1} }$ Definition of Associative Operation $\ds \leadsto \ \$ $\ds e g c^{-1}$ $=$ $\ds c^{-1} g e$ Definition of Inverse Element $\ds \leadsto \ \$ $\ds g c^{-1}$ $=$ $\ds c^{-1} g$ Definition of Identity Element $\ds \leadsto \ \$ $\ds c^{-1}$ $\in$ $\ds \map Z G$ Definition of Center of Group

$\Box$

Therefore:

$\map Z G \le G$

$\blacksquare$

## Proof 2

We have the result Center is Intersection of Centralizers.

That is, $\map Z G$ 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 is Subgroup, $\map Z G$ is also a subgroup of $G$.

$\blacksquare$