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Okay, one last question this semester--pretty stressed out, and can't really figure it out.

We need to show that if $G$ is a doubly transitive subgroup of $S_n$ (that is, for $(x,y)$ distinct, $(u,v)$ distinct, there is some $g$ such that $g(x) = u$ and $g(y) = v$) and if $G$ contains a 3-cycle, then $G = A_n$ or $G = S_n$.

Really I'm just looking for hints in the right direction. Right now I have the intuition that I can pick any two elements and map them anywhere by the doubly-transitive property. And I think we can use that to produce all the 3-cycles, given that the group contains a 3-cycle. And that means we have at least $A_n$. This seems to be a bit of an ad-hoc way of doing it.

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    @flury: Perhaps you can post your solution (as an answer) that the group contains all 3-cycles; then it can be checked for you, *and* the question won't go without an answer.2011-12-14

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So to show that it contains all three-cycles, we use the doubly transitive property. We know that $G$ contains some three-cycle, call it $(x\ y\ z)$. So by the doubly transitive property, there is some $g$ such that $g(x) = u$, $g(y) = v$. So, $G$ must therefore contain $(u\ v\ z)$. Furthermore there is some g'\in G such that g'(u) = u and g'(z) = w, so $G$ contains the three cycle $(u\ v\ w)$ for any arbitrary $u$, $v$, and $w$. Thus, $G$ must contain at least all three cycles, so $A_n\subseteq G$.

If it contains more than that, then it contains some odd permutation $\sigma$, and we can use that to produce the rest of $S_n$.

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    Oh, one more thing: if $f=d$, then just compute $sr^{-1}$.2011-12-14