How to show that the set $Z(G)=\{z \in G : \forall g \in G, z * g = g * z\}$ is a subgroup of $(G, *)$? I have shown that the neutral element is in $Z(G)$. I have also shown that the law is closed in $Z(G)$. 
However, I'm not sure how to prove that $\forall x \in Z(G), \exists x' \in Z(G)$ such that $x * x' = x' * x = e$, where $e$ is the neutral element.
I would like to post what I've tried so far, but everything I think of immediately leads to a dead end. How do I go about solving this? Also, we don't know if $G$ is a commutative group. Thank you.

If $x\in Z(G)$, there is a $x'\in G$ such that $xx'=x'x=e$. So, the problem is to prove that $x'\in Z(G)$. Take $g\in G$. Now, take $g'\in G$ such that $gg'=g'g=e$. Then $xg'=g'x$ (since $x\in Z(G)$), but it follows from this that $gx'=x'g$, as we wanted to prove.
