Center of Group is Normal Subgroup

Theorem
The center of any group $G$ is a normal subgroup of $G$ which is abelian.

Proof of Subgroup

 * 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.


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


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

Therefore, by the Two-step Subgroup Test, $Z \left({G}\right) \le G$.

Alternatively, it can be noted 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 $G$

Thus from Intersection of Subgroups, it must therefore itself be a subgroup of $G$.

Proof of its Abelian Nature

 * The fact that $Z \left({G}\right)$ is abelian follows from the fact that all elements of $Z \left({G}\right)$ commute with all elements of $G$.

Therefore all elements of $Z \left({G}\right)$ commute with all elements of $Z \left({G}\right)$.

Therefore $Z \left({G}\right)$ is abelian.

Proof of its Normality
Since $g x = x g$ for each $g \in G$ and $x \in Z \left({G}\right)$, we have $g Z \left({G}\right) = Z \left({G}\right)g$.

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

Alternatively:
 * $\forall a \in G: x \in Z \left({G}\right)^a \iff a x a^{-1} = x a a^{-1} = x \in Z \left({G}\right)$

Therefore:
 * $\forall a \in G: Z \left({G}\right)^a = Z \left({G}\right)$

and $Z \left({G}\right)$ is a normal subgroup of $G$.