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.


2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – 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.


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