# 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.

• I apologise for ''digging up the dead'' almost 2 years after the original posting of the answer, however it is not clear (to me at least) why establishing the inequality involving the number of classes for the quotient $H$ suffices in order to prove the original claim concerning group $G$. Any clarifications would be appreciated.
– ΑΘΩ
Aug 21, 2020 at 2:44

The argument presented by user8268 is not complete since it does not establish the connection between the lower bound of conjugacy classes on the quotient $$G/\mathrm{Z}(G)$$ and the original number of conjugacy classes of $$G$$. Let me present a very elementary approach to the problem. For arbitrary group $$F$$ I choose to denote the set of all conjugacy classes in $$F$$ by $$\mathscr{Cnj}(F)$$ and its cardinality -- the number of all conjugacy classes -- by $$\mathrm{c}(F)=|\mathscr{Cnj}(F)|$$. For $$x, y \in F$$ I shall write $${}^{x}y=xyx^{-1}$$.

Assume that $$G$$ is finite nonabelian and nevertheless that $$\mathrm{c}(G)=|\mathrm{Z}(G)|+1$$, which means that $$\mathscr{Cnj}(G)=\{\{x\}\}_{x \in \mathrm{Z}(G)} \cup \{A\}$$, $$A$$ being the complementary $$G \setminus \mathrm{Z}(G)$$, by hypothesis itself a conjugacy class. Let us fix a certain $$a \in A$$ and introduce $$B=A \setminus \{a\}$$. Let us also set $$m=|B|$$.

$$G$$ acts transitively by conjugation on $$A$$, which means in other words that the orbital map at $$a$$, given by:

\begin{align} \kappa \colon G &\to A \\ \kappa(t)&={}^{t}a \end{align}

is surjective. It is clear that $$\kappa(\mathrm{Z}(G) \cup \{a\})=\{a\}$$, in other words $$\mathrm{Z}(G) \cup \{a\} \subseteq \kappa^{-1}(\{a\})$$ so we must have $$\kappa^{-1}(B)=G \setminus \kappa^{-1}(\{a\}) \subseteq G \setminus (\mathrm{Z}(G) \cup \{a\})=B$$, by virtue of which we infer that $$|\kappa^{-1}(B)| \leqslant m$$. On the other hand, since $$\kappa$$ is surjective it restricts to a surjection between $$\kappa^{-1}(B)$$ and $$B$$, by virtue of which we also infer the reverse inequality $$m \leqslant |\kappa^{-1}(B)|$$.

Thus, $$B$$ has the same finite cardinal as one of its subsets $$\kappa^{-1}(B)$$ and on grounds of elementary finite cardinal theory we deduce that this latter subset is not proper, in other words $$\kappa^{-1}(B)=B$$. As a consequence we derive that $$\kappa^{-1}(\{a\})=G \setminus \kappa^{-1}(B)=G \setminus B=\mathrm{Z}(G) \cup \{a\}$$. It is easy to see on the other hand that $$\kappa^{-1}(\{a\})=\mathrm{C}_G(a)=\mathrm{Z}(G) \cup \{a\}$$.

Let us write $$k=|\mathrm{Z}(G)|$$. The relation just above entails $$|\mathrm{C}_G(a)|=k+1$$ ($$a$$ is of course noncentral since it belongs to the conjugacy class $$A$$ which is disjoint from $$\mathrm{Z}(G)$$) and since clearly $$\mathrm{Z}(G) \leqslant \mathrm{C}_G(a)$$ -- the centre is clearly a subgroup in the centraliser of any subset -- we must have $$k|k+1$$, which leads to $$k|1$$ and subsequently to $$k=1$$, since $$k$$ is by hypothesis a natural number.

Returning to the original assumption $$\mathrm{c}(G)=|\mathrm{Z}(G)|+1$$, we gather that $$\mathrm{c}(G)=2$$ and we appeal to the (expectedly) well-known and easy to prove fact that any finite group exhibiting only two conjugacy classes is cyclic of order $$2$$ and at any rate abelian. Since $$G$$ is by hypothesis prohibited from being abelian, our assumption is clearly seen to have lead to a contradiction.