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I was trying to solve the following problem : Determine all the finite groups having exactly 3 conjugacy classes.

My attempt:

Among all abelian groups $(\Bbb Z_3,+)$ only satisfies the property , because all other abelian groups of order $n$ , ($n \neq 3$) are having $n$ elements in the center of the group and hence there are exactly $n$ conjugacy classes.

In case of non-abelian groups, if we look at the permutation groups it is again clear that $S_3$ is the only permuatation group having the property because collection of conjugacy classes in permutation groups have one-to-one correspondence with the collection of distinct cycle types.

And in case of Dihedral groups it clearly follows that $D_3$ has the above property and in fact $D_3 \cong S_3$ . The Quarternion group also does not have the property.

So my precise questions are :

$(i)$ Now by considering semi-direct products of these non-abelian groups can the argument be extended?

$(ii)$ Apart from the Permutation groups, about all other non-abelian groups my attempts are mere observations, so how to give a rigorous argument.

I would like to mention that I've gone through this question . It deals with infinte periodic group , but my question is for all infinite groups :

$(iii)$ Is there any infinite group having exactly 3 conjugacy classes .

I have also seen this question and this one as well. They only discussed about finite groups and I was not quite clear with the answers over there.

Hence, trying to generalize the problem :

$(iv)$ For general $n \in \Bbb N$ , how to determine all groups (both finite and infinite) having exactly $n$ conjugacy classes ?

And about the last question I only can state echoing the arguments for $n=3$ , there is exactly one finite abelian group having the property.

Thanks in advance for help.

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    Given $n$ you can bound the number of finite groups with exactly $n$ conjugation classes, I suspect for $n=3$ the bound must be easily computable. For general $n$, I think the answer is "it's impossible to answer it with a general $n$" because it would amount to classifying all finite groups (each finite group falls into one of these families for a given $n$). – Maxime Ramzi Feb 24 '18 at 11:21
  • @lisyarus please read my entire question carefully. –  Feb 24 '18 at 15:40

1 Answers1

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I will answer the original question about $3$ conjugacy classes. Given any $g\in G$, its conjugacy class has size $|G/Z_G(g)|$, where $Z_G(g):=\{h\in G:hg=gh\}$ is the centralizer of $g$. Thus, if $G$ only has three conjugacy classes, represented by $1,g_1,g_2$, then $$|G|~=1+|G|/|Z_G(g_1)|+|G|/|Z_G(g_2)|,$$ i.e., $$1=\frac1{|G|}+\frac1{|Z_G(g_1)|}+\frac1{|Z_G(g_2)|}.$$ At this point, this is a number theory question. Let $|G|=n$ and $|Z_G(g_i)|=n_i$, where $n_i$ are divisors of $n$, and without loss we assume $n>n_1\ge n_2$. Now, we have: $$1=\frac1{n}+\frac1{n_1}+\frac1{n_2}\le\frac3{n_2},$$ so $n_2=2,3$. When $n_2=3$, we have $2/3=1/n+1/n_1$, so in fact $n=n_1=3$. Here, since $Z_G(g_1)=Z_G(g_2)=G$, the group $G$ is abelian, and hence $G=C_3$, as you observe in the question.

Now, if $n_2=2$, then we have $1/2=1/n+1/n_1\le 2/n_1$, so $n_1\le 4$. Thus $n_1=3$ or $n_1=4$. When $n_1=4$ then $G$ has order $n=4$, which must be abelian, a contradiction. If $n_1=3$, then $n=6$, and as you observe, $G=S_3$ here (the only order $6$ groups are $C_6$ and $S_3$).

Note that this indicates a general strategy for answering the question. Suppose your group has $k$ conjugacy classes, and label their representatives as $1,g_1,\dots,g_{k-1}$. Letting $n_i:=|Z_G(g_i)|$ and $n:=|G|$, we have: $$1=\frac1n+\frac1{n_1}+\cdots+\frac1{n_{k-1}}.$$ This equation only has finitely many solutions, and you need to check which ones actually give a $G$ with exactly $k$ conjugacy classes.

Kenta S
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