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Let G be generated by two cycles. One ihas order $n$ and moves the points $\{1,\ldots,n\}$ and the other has order $n-1$ and moves the points $\{1,\ldots,n-1\}$. I found following examples: $$ \begin{array}{ccl} n = 3 &:& S_3 \\ n = 4 &:& S_4 \\ n = 5 &:& S_5,\ C_5 \rtimes C_4\\ n = 6 &:& S_5,\ S_6 \\ n = 7 &:& S_7,\ C_7 \rtimes C_6\\ n = 8 &:& S_8,\ \operatorname{PSL}(3,2) \rtimes C_2\\ n = 9 &:& S_9 \\ n = 10 &:& S_{10} \\ n = 11 &:& S_{11},\ C_{11} \rtimes C_{10} \\ \end{array} $$

Is there a way to predict the structure of these groups for arbitrary $n$?

Edit Changed for $n=6 :S_5$ into $PGL(2,5)$ in accordance with the article mentioned in the answer from @Geoff Robinson.

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With the particular choice $\alpha = (12 \ldots n)$ and $\beta = (12\ldots n-1)$ you will always get $\langle \alpha, \beta \rangle = S_{n}$ as one of the possibilities.This because $\langle \alpha, \beta \rangle = \langle \alpha, \alpha \beta^{-1}\rangle = \langle (12 \ldots n), ( n-1 n) \rangle.$ It is a standard fact, proved in many texts, that $S_{n} = \langle (12 \ldots n), (12) \rangle,$ and it is just a matter of relabelling letters to see that $S_{n} = \langle (12 \ldots n), ( n-1 n) \rangle.$ Equivalently, conjugate $\alpha$ and $\alpha \beta^{-1}$ by the permutation $\sigma$ which interchanges $j$ and $n+1-j$ for $1 \leq j \leq n$ ( and fixes $\frac{n+1}{2}$ if $n$ is odd).

Later edit: Since your group contains an odd permutation and is doubly transitive, I think that the possibilities (other than $S_{n}$) are included in the groups listed in Theorem 1.2 of https://arxiv.org/pdf/1209.5169.pdf ( by G.A. Jones). Note that doubly transitive groups are primitive ( though primitive groups are not always doubly transitive). Note that it is extremely rare in the examples given in this paper that the group contains both an $n$-cycle and an $n-1$-cycle.

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  • $\begingroup$ I knew that already, but I had to include them to be complete, the main point of the question are the other groups. $\endgroup$ – Marc Bogaerts Apr 17 '17 at 19:31
  • $\begingroup$ OK. The groups you get will always be doubly transitive, and doubly transitive groups are reasonably well understood now ( but a full understanding requires the classification of finite simple groups). $\endgroup$ – Geoff Robinson Apr 17 '17 at 19:38
  • $\begingroup$ The examples I gave are exhaustive for the given $n$. The cases of the form $C_n \rtimes C_{n-1}$ can also be explained away for $n$ prime because of the structure of the automorphism group of $C_n$. Remain the special cases for $n=6$ and $n = 8$ and who knows for other values of $n$. $\endgroup$ – Marc Bogaerts Apr 17 '17 at 19:41
  • $\begingroup$ Indeed, I was inspired by this question. I had an answer for the $S_n$ case but had to delete it because I realized there were other groups falling under this description. $\endgroup$ – Marc Bogaerts Apr 17 '17 at 19:48
  • $\begingroup$ I have edited my answer to include some more information. $\endgroup$ – Geoff Robinson Apr 17 '17 at 19:49

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