In connection with the research discussion about graphs and groups, I began to wonder which finite groups have the property that any two elements of the same order are conjugate. I thought about this, and got a certain distance, and then other jobs crowded it out of my thoughts.

I had become convinced that there are only three such groups, the symmetric groups of degrees 1, 2 and 3, and had got part way to a proof. In the end, not having time to go back to it, I resorted to a well-known search engine. With a little further research I came up with the answer, and with a small problem, which I want to pose.

In 1933, G. A. Miller published a paper in the *Proceedings of the National Academy of Sciences* of the USA on the problem. (His name is one which has come up a number of times in our researches about graphs on groups.) He showed, in brief, that in a group with this property, the derived group has index at most 2; moreover, if it has index 2, then the group is the symmetric group of degree 2 or 3. He couldn’t deal with the remaining case (though he did reduce it to considering non-abelian simple groups), and the earliest solution of the problem I found was by Patrick Fitzpatrick, in the *Proceedings of the Royal Irish Academy* in 1985; he used the Classification of Finite Simple Groups to show that the only such group is the trivial group. This was proved again by Walter Feit and Gary Seitz in 1988. (As a student of Peter Neumann, I know better than to say that they reproved the result – he would certainly have reproved me for saying so!)

So the problem I wish to pose is to give a proof of this simple result which does not depend on CFSG.

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About Peter Cameron

I count all the things that need to be counted.

Pasha Zusmaonvich tells me that this is Problem 7.48 in the Kourovka Notebook; also, the paper of Feit and Seitz is in 1989.