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.

  • 1
    $\begingroup$ 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 $n$). $\endgroup$ – Max Feb 24 '18 at 11:21
  • $\begingroup$ @lisyarus please read my entire question carefully. $\endgroup$ – reflexive Feb 24 '18 at 15:40

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