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"Let $G$ be a transitive group of permutations on a given set of letters. Let a new fixed letter be adjoined to every permutation of $G$. Then a transitive group $H$ of permutations on the combined set of letters is called a transitive extension of $G$ if it contains $G$ as the largest subgroup fixing the new letter."

... T. C. Holyoke, Transitive extensions of dihedral groups, Math. Zeitschr. Bd. 60, S. 79/80 (1954)

In this paper, Holyoke determined that the transitive extensions of a dihedral group $D_n$ acting in the usual way on $n$ points, are limited to the extensions: $S_3$ (of $D_2$), $S_4$ (of $D_3$) and $PSL(2, 5)$ of $D_5$. My question is, what are all the transitive extensions of cyclic groups?

EDIT: Here is a modern equivalent formulation of the question: "What are the $2$-transitive groups with a cyclic point-stabiliser?"

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  • $\begingroup$ This formulation was already, 1954, somewhat old-fashioned, if I compare to articles of the same time by Higman or Ph. Hall... would be nice to translate into modern mathematical language. $\endgroup$
    – YCor
    Jan 29, 2018 at 4:54

2 Answers 2

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Updated: In the finite case (as in the reference), the only examples are $\mathrm{AGL}(1,q)$ for $q$ a prime power.

This follows, for example, by "Lucchini, Mainardis, Stellmacher, Transitive permutation groups with cyclic point stabilizers of maximum order. Geom. Dedicata 100 (2003), 117–121."

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    $\begingroup$ The proof in the link you give is extreme overkill and uses the classification of finite simple groups. The result here can proved much more easily. As you noted yourself, it is a Frobenius group, so has a regular normal subgroup, which must be elementary abelian because all of its nontrivial elements are conjugate. Now you can use the result that representations of abelian groups in coprime characteristic have dimension one over an extension field to set up the isomorphism wtih ${\rm AGL}(1,q)$. $\endgroup$
    – Derek Holt
    Jan 29, 2018 at 9:47
  • $\begingroup$ I'd be happy if you started your post by reformulating the question in modern language...! $\endgroup$
    – YCor
    Feb 6, 2018 at 23:08
  • $\begingroup$ Yes, but my initial request to the OP was unsuccessful. $\endgroup$
    – YCor
    Feb 7, 2018 at 4:04
  • $\begingroup$ Done. (Also cleaned up answer, and deleted some now irrelevant comments.) $\endgroup$
    – verret
    Feb 8, 2018 at 0:24
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Let the cyclic group on question have order $k-1.$ In the event that $k$ is a prime power, the multiplicative group $F_k^*$ of the field of order $k$ is cyclic on the non-zero elements and is the stabilizer of $0$ in the affine group of all non-constant transformations $f(x)=mx+b$ which acts transitively on the $k$ elements of the field.

I suspect that those are the only examples.

In general (if I understand correctly) what you want is exactly a transitive subgroup of $S_k$ which has order $(k-1)k$ and contains a $k-1$ cycle.

The order alone might be sufficient. For $k=6$ there are no order $30$ subgroups of $S_6.$ There are also no order $90$ subgroups of $S_{10}.$

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