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 $$G$$.