# 2-transitive subgroup of $S_n$ containing a 3-cycle

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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Nothing inherently wrong with being ad hoc, so long as it works. The proof that $A_n$ is generated by 3-cycles (or that $A_n$ is simple for $n\geq 5$) is pretty ad hoc itself... –  Arturo Magidin Dec 13 '11 at 21:16
Yeah, I guess I am okay with what I've got. I need to show that if we have more than $A_n$ (so at least one odd permutation) then we have $S_n$ –  flury Dec 13 '11 at 21:17
The problem asks you to show that the group will be either $A_n$ or $S_n$. If you can show that the group will contain all $3$ cycles, then you are done: it must contain $A_n$, and since $A_n$ is maximal in $S_n$ with $A_n\subseteq G\subseteq S_n$, you know $A_n=G$ or $G=S_n$. –  Arturo Magidin Dec 13 '11 at 21:20
@flury That second part should be clear, no? If you have an odd $\sigma$ and you have all of $A_n$, then you get the coset $\sigma A_n$ and that's everything. So maybe try to write up a proof that at least $A_n$ is in $G$? –  Dylan Moreland Dec 13 '11 at 21:21
How are you showing it contains all 3 cycles? –  user641 Dec 14 '11 at 0:34

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 $S_n$.
It seems like you're conjugating by $g$ to get $(u\ v\ z)$. How do you know that $g$ doesn't move $z$ somewhere? –  Dylan Moreland Dec 14 '11 at 6:20
As Dylan mentioned, this is not correct. Try the following: pick a $g\in G$ that moves $x$ to $a$ and $z$ to $b$, giving the 3-cycle $r=(adb)$. Now pick an $h$ moving $x$ to $a$ and $z$ to $c$, giving a 3-cycle $s=(afc)$. Now compute $[r,s]$. –  user641 Dec 14 '11 at 17:39
Sorry, I should mention that if either $d=c$ or $f=b$, you're already done, so you can assume that neither happens. –  user641 Dec 14 '11 at 17:41
Oh, one more thing: if $f=d$, then just compute $sr^{-1}$. –  user641 Dec 14 '11 at 17:45