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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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    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.2012-01-04

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