# Unicyclic groups

When working on large finite field versions of conjectures like Bunyakowski’s or Schinzel’s or Bateman-Horn’s, one small point that always somewhat tickled me was that the answer, which follows from the computation of some Galois group, seems to be weaker than this actual computation. For instance, to get the “right” answer in Bunyakowski’s Conjecture, one computes a Galois group $G$ that is in a natural way a subgroup of the symmetric group on $n$ letters, and it is enough to show that the proportion of $n$-cycles in $G$ is exactly $1/n$. It is of course sufficient to show that $G$ is the full symmetric group (which contains the right number of $n$-cycles), but a priori, this is a weaker condition.

Although I mentioned this briefly in my talk in the recent conference in honor of A. Lubotzky’s 60th birthday (where I was greatly honored to be invited!), it’s only last week or so that I somehow finally did the obvious thing, namely some experiments with Magma to see if the property of having proportion $1/n$ of $n$-cycles is widespread.

When doing this, the first thing to realize (which I only did when P.P. Palfy pointed it out during that conference…) is that this condition, for a subgroup $G$ of $\mathrm{Sym}_n$, is equivalent with $G$ containing a unique conjugacy class of $n$-cycles (simply because the centralizer of an $n$-cycle, either in $G$ or in the bigger symmetric group, is the cyclic subgroup of order $n$ that it generates). So we can coin a solid decent name for these groups: we call them the unicyclic permutation groups.

Magma has a database of all transitive permutation groups of degree up to $31$ (even $32$ if one installs an extra specific database). Experimentation shows that there often exist unicyclic groups that are not the symmetric group. For instance, for degree $20$, there exist (up to isomorphism) $1117$ transitive permutation groups, and $332$ of them contain at least one $20$-cycle, and $35$ of them are unicyclic.

However, the same experiments show that if we restrict our attention to primitive subgroups, then the situation is very different: either there is only the symmetric group $\mathrm{Sym}_n$, or there are two unicyclic groups. Amusingly the second case occurs if and only if $n$ is prime (an amusing primality test…)

One can indeed prove that these facts (which I did first experimentally notice) are correct, but it is not cheap. More precisely, from the Classification of Finite Simple Groups, Feit and later Jones (see for instance this paper which has a slightly more general result) deduced a full classification of primitive permutation groups that contain an $n$-cycle. It is restrictive enough that one can then relatively easily exclude all groups except the symmetric groups and, when $n$ is prime, the group of transformations $x\mapsto ax+b$ of the finite field $\mathbf{F}_n$.

That’s a nice result, but is it relevant for our original motivation? I think so, because (as Will Sawin pointed out), one can prove in considerable generality that the Galois group in (say) the Bunyakowski problem is primitive, before or without computing it exactly, because it suffices to prove that it is $2$-transitive, and this is accessible to the diophantine interpretation using Chebotarev’s Density Theorem, as in Entin’s approach. (Indeed, Entin goes on to check that the Galois group is often $7$-transitive or so, and from the Classification of Finite Simple Groups, deduces that the group contains the alternating group). Hence, if the degree $n$ is not prime, granted that one first applies Entin’s method to check primitivity, having Galois group $\mathrm{Sym}_n$ is equivalent to the natural version of Bunyakowski’s conjecture in the large finite field limit…
If $n$ is prime, well one can say that if $G$ is not solvable, then it must be the symmetric group (assuming $n$ is at least $5$…), although that is not as good — is there a better way to do it?