# Finite Groups with exactly $n$ conjugacy classes $(n=2,3,...)$

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?

• 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. Jul 19, 2011 at 7:50
• Jul 19, 2011 at 9:38
• 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/…
– user9413
Jul 19, 2011 at 11:32
• Another approach for $n=2$, we can also consider the faithfull action of $G$ on the set of all conjugacy classes of $G$ via $g.x^{G}=(gx)^{G}$. Jan 8, 2014 at 14:08

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.

• @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. 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.

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