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.
Best Answer
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.