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Let $n$ be an odd natural number (sufficiently large), and $ 1\leq t,k < n $. Let $A= \{ a_{1},a_{2},\ldots,a_{t} \} $, $B= \{ b_{1},b_ {2},\ldots,b_{n-t} \} $, $C= \{ c_{1},c_{2} \ldots,c_{k} \} $ and $D= \{ d_{1},d_{2},\ldots,d_{n-k} \} $ be subsets of $ \{ 1,2,\ldots,n \} $ such that $A\cup B=C\cup D= \{ 1,2,\ldots,n \} $ and $A\cap B=C\cap D=\phi$ and $A\neq C$ and $A\neq D$. What can we say about the subgroup generated by $(a_{1},a_{2},\ldots,a_{t})(b_{1},b_{2},\ldots,b_{n-t})$ and $(c_{1},c_{2},\ldots,c_{k})(d_{1},d_{2},\ldots,d_{n-k})$. I think that this subgroup is $S_{n}$. Is it true? Maybe there is a simple counterexample but I've not found it.

Clearly, $n$ should be odd. Otherwise, this group is a subgroup of $A_{n}$.

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3 Answers

up vote 5 down vote accepted

Counterexample: Everything in $\langle(123)(456789),(123456)(789)\rangle$ acts cyclically on the set partition $\lbrace1,4,7\rbrace \lbrace2,5,8\rbrace \lbrace 3, 6, 9 \rbrace$. This type of counterexample works for any odd composite $n$.

Thanks to Michael Zieve for pointing out that the following is incorrect:

For prime $p$ the result is true by the O'Nan-Scott Theorem, the classification of maximal subgroups of the symmetric (and alternating) groups.

One of the cases of the O'Nan-Scott classification of maximal subgroups of the symmetric group is allowed: $AGL(1,p)$ acting on the affine line.

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That "one of the cases" should be "at least one of the cases." –  Douglas Zare Feb 22 '13 at 23:48
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Regarding the assertion that the result is true for prime $n$, here are counterexamples for every prime $n>3$. A simple concrete example for $n=5$ is given by $(1,2,3,4)(5)$ and $(1,3,5,4)(2)$, which generate an order-$20$ subgroup of $S_5$. More generally, let $g$ be a generator of the multiplicative group of $\mathbb{F}_n$, and consider the permutations of $\mathbb{F}_n$ given by $x\mapsto gx$ and $x\mapsto gx+1$. These two permutations generate the group of all permutations $x\mapsto ax+b$ with $a,b\in\mathbb{F}_n$ and $a\ne 0$, which has order $n(n-1)$. But each of these two permutations has one fixed point and one $(n-1)$-cycle, and their fixed points are distinct.

Note that the reason $n=3$ is excluded is that this order $n(n-1)$ subgroup of $S_n$ equals $S_n$ when $n=3$.

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I believe that the two permutations $$ (1\ 2\ 3)(4\ 7\ 5\ 8\ 6\ 9) \quad\text{and}\quad (4\ 5\ 6)(1\ 7\ 2\ 8\ 3\ 9) $$ generate a subgroup of $S_9$ of order only $162$.

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Magma confirms this. –  Denis Chaperon de Lauzières Feb 21 '13 at 21:39
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