TITLE: What are those groups generated by two cycles?
QUESTION [9 upvotes]: 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.

REPLY [5 votes]: 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.