I am looking to classify (up to isomorphism) those finite groups $G$ with exactly 2 conjugacy classes.

If $G$ is abelian, then each element forms its own conjugacy class, so only the cyclic group of order 2 works here.

If $G$ is not abelian, I am less sure what is going on. The center $Z(G)$ is trivial since each of it's elements also form their own conjugacy class. Now assume $G-\{1_G\}$ is the other conjugacy class.

The Class Equation says $|G|=|Z(G)|+\sum [G:C_G(x)]$ where the sum is taken over representatives from the conjugacy classes (not counting the singleton ones from the center). (Here $C_G(x)=\{g\in G~|~gx=xg\}$ is the centralizer of $x$ in $G$.)

For us this simplifies to $|G|-1=[G:C_G(x)]$ for any $x\in G-\{1_G\}$. Therefore $|C_G(x)|=\frac{|G|}{|G|-1}$ is an integer. But this can only happen when $|G|=2$ which we have already covered. So does this mean up to isomorphism there is only one group with 2 conjugacy classes?

If so, how would the argument change if we allowed 3 conjugacy classes?

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    $\begingroup$ Hint: there are exactly two isomorphism types of finite groups with exactly three conjugacy classes. You can show this using similar arguments to the case of two classes. $\endgroup$
    – Derek Holt
    Jul 19, 2011 at 7:50
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    $\begingroup$ See also math.stackexchange.com/questions/46981/… $\endgroup$ Jul 19, 2011 at 9:38
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    $\begingroup$ I don't know whether to pose this as an answer or a comment, but certainly there is something very interesting in this link: groupprops.subwiki.org/wiki/… $\endgroup$
    – user9413
    Jul 19, 2011 at 11:32
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    $\begingroup$ Another approach for $n=2$, we can also consider the faithfull action of $G$ on the set of all conjugacy classes of $G$ via $g.x^{G}=(gx)^{G}$. $\endgroup$ Jan 8, 2014 at 14:08

4 Answers 4


Nice question! The $n = 3$ case is fun and I think small values of $n$ are going to make very good exercises so I encourage you to work on them yourself, but if you really want to know a solution....

If $H_1, H_2$ denote the stabilizers of the non-identity conjugacy classes with $|H_1| \le |H_2|$, then the class equation reads $|G| = 1 + \frac{|G|}{|H_1|} + \frac{|G|}{|H_2|}$, or

$$1 = \frac{1}{|G|} + \frac{1}{|H_1|} + \frac{1}{|H_2|}.$$

The reason this is useful is that if either $|G|$ or $|H_1|$ gets too big, then the terms on the RHS become too small to sum to $1$. Since we know that $|G| \ge 3$, it follows that we must have $|H_1| \le 3$; otherwise, $\frac{1}{|H_1|} + \frac{1}{|H_2|} \le \frac{1}{2}$ and the terms can't sum to $1$.

Now, if $|H_1| = 2$ then $|H_2| \ge 3$, hence $|G| \le 6$. Since every group of order $4$ is abelian we can only have $|G| = 6, |H_2| = 3$. The unique nonabelian group of order $6$ is $S_3$, which indeed has $3$ conjugacy classes as desired.

If $|H_1| = 3$, then $|G| \ge 3$ implies $|H_2| \le 3$, hence $|H_2| = |G| = 3$ and $G = C_3$.


You may want to know that in case of infinite groups the situation is entirely different. In 1949 Graham Higman, Bernard H. Neumann and Hanna Neumann wrote a now world-famous paper called Embedding Theorems for Groups. It embeds a given group $G$ into another group $\tilde{G}$, in such a way that two given isomorphic subgroups of $G$ are conjugate (through a given isomorphism) in $\tilde{G}$. Hence one can embed any countable group in a group with the property that any two elements of equal order are conjugate. So, using that result, you need only begin with your favourite (non-trivial) torsion free group, and you get an infinite group with only two conjugacy classes! The proof of this HNN-extension construction uses the idea of taking an ascending union. At each step, you can use the HNN extension construction to embed $G_k$ in a group $G_{k+1}$, in which any two elements of $G_k$ are conjugate provided only that their orders are equal (it may be, however, that two elements of $G_{k+1}$ with infinite orders are not conjugate in $G_{k+1}$ itself). After forming the union of the chain ${G_k}$ of groups, any two members with the same order belong to some $G_n$, and they are then conjugate in $G_{n+1}$.


Please see $\text{Bounding size by Number of Conjugacy Classes}$ in the below Keith Conrad's article.

  • $\begingroup$ @KCd: Thanks for letting me know . I will delete my comment . I have always been interested in how to control (by various means) the number of conjugacy classes of a finite group. $\endgroup$ Sep 20, 2020 at 9:32

The problem of classifyng finite groups by the number of conjugacy classes is classical, and as far as I can tell (group theory is not my field), hard. In this paper, the authors classify all finite groups with at most $11$ conjugacy classes.

  • $\begingroup$ If you look at Nicky Hekster's link in the comments to the question you'll find references beyond 11. $\endgroup$ Jul 19, 2011 at 12:53

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