TITLE: Let $G$ denote an arbitrary group. Prove: The center of any group $G$ is a normal subgroup of $G$
QUESTION [5 upvotes]: Let $G$ denote an arbitrary group. Prove: The center of any group $G$ is a normal subgroup of $G$

Let $G$ be a group and $C$ the center, i.e., for any $a \in C$ and any $x \in G$, $xa=ax$.
So, $a = xax^{-1}$. Thus $xax^{-1} \in C$. So, $C$ is normal.
Is this proof correct? It seemed too trivial for me to think so.
Something tells me I need to show that for any $y \in G$, $xax^{-1}y=yxax^{-1}$. Would this be correct? But this would really be the same as above.

REPLY [2 votes]: Let $G$ be a group and let $Z(G)$ be the center of $G$. 
If $x \in Z(G)$, then $xy=yx$ $\ \forall y \in G$. Thus $yxy^{-1}=x \in Z(G)$.
So, $Z(G) \vartriangleleft G$.