I have seen the question on groups with two conjugacy classes, and I proved to myself that such a group must be torsion-free (if it isn't the cyclic group of order 2), but what about a group with three conjugacy classes? If it is finite, it must be C3 or S3, but what if it is infinite?

I also proved that there is no infinite group with three conjugacy classes and exponent 3 (first prove that two conjugate elements commute, then prove that the third class is the inverses of the elements in the first class, and from those two facts it follows that the group is abelian). No other prime exponent works, so we have to move on to the case of exponents with two factors.

Is there an infinite group with only three conjugacy classes and finite exponent?


In general, there is no infinite periodic group with only three conjugacy classes.

Assume that $G$ is a counter example. Then either $G$ has exactely one infinite conjugacy class or two of them.

It is known that the elements having finite conjugacy classes form a subgroup $F=FC(G)$ of $G$, called the FC-center of $G$.

In the case where $G$ has exactely one infinite conjugacy class, any two non-trivial elements in $G/F$ are conjugates, so $G/F$ is simple of prime exponent $p$ as it is periodic.

Now for any non-trivial element $x \in G/F$, $1, x, ..., x^{p-1}$ lie in different conjugacy classes. Indeed, if $x^i$ and $x^j$ are conjugates, we can assume that $x$ and $x^k$ are conjugates for some integer $k \neq 1 \mod p$. We have $x^g=x^k$, for some $g \in G/F$, thus $x=x^{g^p}=x^{k^p}$. It follows that the order of $k$ in the multiplicative group of the field of integer modulo $p$ is equal to $p$, a contradiction.

Thus we may assume that $G/F$ has exponent $2$, so it is abelian. By our condition on the conjugacy classes, it follows that $G/F$ has order $2$. And as $F$ is a union of two finite conjugacy classes, we conclude that $G$ is finite.

Now we have to assume that $G$ has exactely two infinite conjugacy classes. Our claim now follows from the following result due to Izosov (see also "On groups with two infinite conjugacy classes" by M. Herzog, P. Longobardi and M. Maj, in Ischia Group Theory 2006)

If $G$ is a group with exactely two infinite conjugacy classes and $G/F$ is periodic (where $F$ denotes the FC-center of $G$), then $2 \leq |G:F| \leq 3$. (Note that in our case $F=1$).


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