Prove: If $G$ is a nonabelian group with $[G : Z(G)] = n$, then every conjugacy class of $G$ has strictly fewer than $n$ elements.
My approach so far:
Observe that $Z(G) \leq C_G(a) = G_a \leq G$. So we know $$[G:Z(G)] = [G:G_a][G_a:Z(G)] = \vert C_a \vert [G_a : Z(G)]$$ and thus we can conclude that $\vert C_a \vert = [G:G_a] \mid [G:Z(G)] = n$.
Now, suppose to the contrary that $\vert C_a \vert = n$. [Try to contradict $G$ being nonabelian.]
I'm having trouble finding a contradiction to the case of $n$. Any hints, approaches? Thanks!