# Minimum number of conjugacy classes of a finite non-abelian group

I am trying to answer the following: Let $$G$$ be a finite non-abelian group. Show that $$k(G) > |Z(G)| + 1$$, where $$k(G)$$ is the number of conjugacy classes of $$G$$.

I can see that each element in $$Z(G)$$ lies in a conjugacy class on its own. Also $$G$$ must have an element not in its centre because it is non-abelian and this element must be in another conjugacy class (which must contain at least 2 elements).

This gives $$|Z(G)| + 1$$ conjugacy classes so far. I am not sure how to show that there must be at least 1 more though.

If $$H=G/Z(G)$$, and if you show that $$k(H)>2$$, then you're done. The only way we might have $$k(H)=2$$ is that $$H\setminus\{1\}$$ is a conjugacy class. As the number of elements of a conjugacy class divides the order of the group, we must have $$|H|=2$$, i.e. $$H=C_2=\{1,-1\}$$ (cyclic group with $$2$$ elements).
We thus have a group extension $$1\to Z(G) \to G \to C_2\to1$$. If $$a\in G$$ is mapped to $$-1\in C_2$$ then $$G$$ is generated by $$Z(G)$$ and by $$a$$, and $$a$$ commutes with $$Z(G)$$, so $$G$$ is commutative, giving a contradiction.