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This is the special case I am working on (from this wikipedia page on symmetric groups: https://en.wikipedia.org/wiki/Inverse_Galois_problem#Symmetric_and_alternating_groups). How does one show that if Gal(f(x, s)/Q(s)) acts doubly transitive on the roots of f and if it contains a transposition, then Gal(f(x, s)/Q(s)) must be the full symmetric group Sn. The more general case of this theorem is in the title above. I know how to prove that a transitive subgroup with a transposition of Sp (p prime) must be equal to Sp, but don't know how to prove this for general n integer.

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Let $G \subseteq S_n$ be a group acting doubly transitive on $\{1,\dots,n\}$ and suppose $(1,2) \in G$. Now consider $i,j \in \{1,\dots, n\}$, $i \neq j$. Since $G$ acts doubly transitive, there exists $\sigma \in G$ such that $\sigma(1) = i$ and $\sigma(2) = j$. We have $$(i,j) = \sigma (1,2) \sigma^{-1} \in G.$$ It follows that $G$ contains all transpositions and since $S_n$ is generated by transpositions, we have $G = S_n$.

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  • $\begingroup$ thanks good answer $\endgroup$ – Daniele1234 Aug 23 '17 at 12:32

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