# centralizer of the symmetric group

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 $|X|=n-k$?

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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
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

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.

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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.

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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