Center of Group is Normal Subgroup

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

Proof

 * By the definition of identity, $$eg=ge=g$$ for all $$g \in G$$.

So, $$e \in Z(G)$$, meaning $$Z(G)$$ is nonempty.


 * Suppose $$a,b \in Z(G)$$.

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

$$\forall g \in G: (ab)g=a(bg)=a(gb)=(ag)b=(ga)b=g(ab)$$.

Thus, $$ab \in Z(G)$$.


 * Suppose $$c \in Z(G)$$.

Therefore, by the two-step subgroup test, $$Z(G) \le G$$.

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 which are subgroups of $$G$$, and from Intersection of Subgroups, must therefore itself be a subgroup of $$G$$.


 * 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.

Since $$gx=xg$$ for each $$g \in G$$ and $$x \in Z(G)$$, we have $$gZ(G)=Z(G)g$$. Thus, $$Z(G) \triangleleft G$$.

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 $$G$$.