MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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 $A_{n}$.

share|cite|improve this question
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.

share|cite|improve this answer
That "one of the cases" should be "at least one of the cases." – Douglas Zare Feb 22 '13 at 23:48

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 $n=3$.

share|cite|improve this answer

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 $162$.

share|cite|improve this answer
Magma confirms this. – Denis Chaperon de Lauzières Feb 21 '13 at 21:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.