Center of Group is Normal Subgroup

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

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

Suppose $$c \in Z(G)$$. Then $$cg=gc,$$ $$\forall g \in G$$. So, $$c^{-1}(cg)c^{-1}=c^{-1}(gc)c^{-1},$$ implying $$(c^{-1}c)gc^{-1}=c^{-1}g(cc^{-1}),$$ implying $$egc^{-1}=c^{-1}ge,$$ giving us $$gc^{-1}=c^{-1}g$$. Thus, $$c^{-1} \in Z(G)$$.

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

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