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I know that the symmetric group $S_n$ is generated by $(12)$ and $(2345\dots n)$.

Let $G$ be a transitive subgroup of $S_n$ (transitive with respect to the natural action of $S_n$ on $\{1,2,\dots,n\}$) that contains a transposition and an $(n-1)$-cycle. Prove that $G=S_n$.

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    $\begingroup$ Up to conjugation you can suppose your $(n-1)$-cycle to be a given one, say $(23\ldots n)$. What could the transposition that accompanies this $(n-1)$-cycle be? $\endgroup$ – Marc van Leeuwen Jan 4 '12 at 8:02
  • $\begingroup$ When you say up to coniugation you mean that after a relabelling of 1,2..n I can suppose my (n-1)cycle to be (23..n). Then becouse G is transitive I can suppose my transposition to be (1k)with k GEQ 2:if 2 I conclude,otherwise I can transform (1k)to (12)with (23..n)and conlude using the hypothesis.Is it right? Thanks $\endgroup$ – Steve0078 Jan 4 '12 at 10:44
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    $\begingroup$ Yes, relabelling elements really amounts to conjugation by a permutation. And because $G$ is transitive, the transposition must involve the fixed point $1$ of the cycle (I would not say "can suppose" here). By conjugating by a power of $(23\ldots n)$, which will not change this cycle itself, the transposition can be made to be $(12)$. So you guessed the solution; feel free to answer your own question to provide a nicely formulated answer. $\endgroup$ – Marc van Leeuwen Jan 4 '12 at 10:51

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