Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If $H$ and $D$ are subgroups of $S_n$, the symmetric group on $n$ letter, such that $H\cong S_k$ and $D\cong S_{n-k}$, where $n>k>1$, $n>15$ and $H=C_{S_n}(D)$ and $D=C_{S_n}(H)$, Is it true that $H=Stab_{S_n}(X)$ and $D=Stab_{S_n}(X^c)$, where $X$ is a subset of $\{1,2,\dots,n\}$ such that $|X|=n-k$?

share|improve this question
1  
Is there any specific reason why you assume $n>15$? Is the result wrong for smaller $n$? –  Johannes Ebert Jun 12 '12 at 7:33
    
It looks like you are endowing $\{1,\ldots,n\}$ with the standard action of $S_n$ by permutations. If that is the case, then it is easy to see that the answer is "no", because you can conjugate a pair of standard subgroups nontrivially. If you are just asking if $H$ and $D$ are simultaneously conjugate to the subgroups listed, then the question is more interesting. –  S. Carnahan Jun 12 '12 at 7:41
2  
S. Carnahan: I don't understand your comment. I believe the result is true as stated. –  Derek Holt Jun 12 '12 at 9:24
    
@S. Carnahan: Conjugating the groups $H$ and $D$ by a permutation $\pi$ would just replace $X$ by $\pi(X)$, so the proposed conclusion, the existence of a suitable $X$, would be unaffected. –  Andreas Blass Jun 12 '12 at 10:19
add comment

2 Answers

Yes. One way to see this is to use the fact that the smallest degree faithful permutation representations of $S_n$ for $n \ge 6$ have degrees $n$ (natural action), $2n$ (imprimitive action on cosets of $A_{n-1}$), and $n(n-1)/2$ (action on unordered pairs). I think this was probably proved before the classification of finite simple groups, but I am not sure. A general reference for maximal subgroups of $A_n$ and $S_n$ is

Liebeck, Martin W.; Praeger, Cheryl E.; Saxl, Jan. A classification of the maximal subgroups of the finite alternating and symmetric groups. J. Algebra 111 (1987), no. 2, 365–383.

Let's assume $n \ge 11$ (although I think your result is true for $n \ge 7$) and $k \ge n-k$, so $k \ge 6$. So the only faithful permutation representations of $S_k$ of degree at most $n$ have degree $k$, or $2k$ when $n=2k$.

The imprimitive transitive representation of $S_k$ of degree $2k$ has centralizer of order 2 in $S_{2k}$ (the order of the centralizer in the symmetric group of a transitive permutation group is equal to the number of fixed points of its point-stabilizer), so that cannot arise as your subgroup $H \cong S_k$.

Hence $H$ must have an orbit of length $k$ and act naturally on that orbit. If it had two orbits of length $k$ (with $n=2k$), then its centralizer in $S_n$ would have order 2, so it has a unique such orbit. So your subgroup $D$ isomorphic to $S_{n-k}$, which centralizes $H$, must fix that orbit, and hence fix it pointwise, and so $D$ must act naturally on the remaining points. Hence $H$ fixes all remaining points, and the result follows.

share|improve this answer
add comment

Johannes, I need $n>15$ in my work, however Magma software show it is not true for $n=6$, but for $n=7,8,9$ is true.

share|improve this answer
2  
If you register an account, you can edit your question instead of adding answers where they don't belong. –  S. Carnahan Jun 12 '12 at 8:32
    
One would expect the result to fail for $n=6$, because $S_6$ has outer automorphisms. Applying an outer automorphism to an $H$ and $D$ of your desired form, you'd get new $H$ and $D$ not of that form. (In other words, an outer automorphism works as well as a conjugation in S. Carnahan's comment to the question, and, unlike a conjugation, it does not preserve the desired conclusion.) It would be interesting if there were, for $n=6$, counterexamples other than those given by outer automorphisms. –  Andreas Blass Jun 12 '12 at 10:23
add comment

Your Answer

 
discard

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.