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