Proof envolving center of a group and conjugacy classes

Question

I need to prove that every group $G$ of order 9 is abelian.
So, this is what I've done so far:
There are three options for $[G:Z(G)]$:
$[G:Z(G)]=1$, in this case we are done, since $G=Z(G)$ and that means $G$ is abelian. $[G:Z(G)]=3$, in this case, we have that $G/Z(G)$ is cyclic, and that implies that $G$ is abelian (I proved it earlier).
$[G:Z(G)]=9$, since this means that $Z(G)=\left\{e\right\}$, we have that: $|G|=|Z(G)|+\sum_{x\notin Z(G)}|[x]|$ (while $[x]$ means the conjugacy class represented by $x$) and then $\sum_{x\notin Z(G)}|[x]|=8$, but I couldn't find the contradiction here.
Any help would be appreciated.

Answer 1

Tobias's comment already solves the question: what are the possibilities if

$$\sum_{x\notin Z(G)}|[x]|=8\;,\,\,|[x]|\mid 9\;\;\text{and}\;\;|[x]|>1\;?$$

Of course, it is impossible so it can't be $\;|Z(G)|=1\;$ , and thus you're done.