Suppose $G=S_n$ is the permutation group in $n$ letters and $7\leq n\leq 10$. Also consider the subgroups $H_1,H_2$ such that $H_1$ is isomorphic with $S_k$ and $H_2$ is isomorphic with $S_{n-k}$ and every element of $H_1$ commutes with every element of $H_2$. Is it true that $H_1,H_2$ are the stabilizers of $n-k$ and $k$ letters respectively?
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1$\begingroup$ For $k=1,2,3,\lfloor n/2\rfloor$, you should consider all possible actions of $S_{n-k}$ (or $S_k$, if easier) on $n$ elements, compute their centralizers and check what's going on. What have you tried so far? $\endgroup$– YCorCommented Sep 2, 2016 at 9:17
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2$\begingroup$ $S_5$ has a transitive action in $S_6$, and the centralizer is $S_1$ (i.e. the trivial group). So there's a counter-example. There might be others for small $n$. But I'd expect the statement is true for $n$ large enough -- use the fact that the centralizer of a transitive group is semi-regular. $\endgroup$– Nick GillCommented Sep 2, 2016 at 9:25
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3$\begingroup$ By the way, I don't understand why there is a vote to close. The question seems perfectly reasonable to me. $\endgroup$– Nick GillCommented Sep 2, 2016 at 9:27
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2$\begingroup$ @YCor, Fair enough. I think the question is OK here. But MSE would be fine too. $\endgroup$– Nick GillCommented Sep 2, 2016 at 9:36
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1$\begingroup$ A similar idea to Nick's gives a counterexample with $k=2$ or $k=3$. $S_6$ has an outer automorphism (this sends the stabilizer of a point to the transitive $S_5$ Nick mentioned). So consider stabilizers of $6-k$ and $k$ letters in $S_6$ (for $k=2$ and $k=3$) and then take their images under the outer automorphism. $\endgroup$– benblumsmithCommented Sep 2, 2016 at 12:14
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Pick a prime $p$ in the range $\frac{n}{2}<p\leq n-k$. Then $S_{n-k}$ contains $(p-1)!\binom{n-k}{p}$ elements of order $p$, and elements of order $p$ in $S_n$ are $p$-cycles. There are at most $(p-1)!$ cycles of length $p$ in $H_2$, which move the same $p$ points, hence $H_1$ fixes the union of $\geq\binom{n-k}{p}$ sets of size $p$ pointwise, and this union has $\geq n-k$ elements. Hence $H_1$ is the stabilizer of a set of $\geq n-k$ points, and $>$ is clearly impossible.
If there is no prime in this range, things get a bit more complicated. Suppose $p<q$ are primes with $p+q\leq n-k$. Then an element of order $pq$ in $H_2$ consists of $\alpha$ cycles of length $p$ and $\beta$ cycles of length $q$. If $\alpha=\beta=1$, you can argue as above. Otherwise the centralizer of this element in $H_1$ has order $\leq \alpha! \beta! p^{\alpha-1} q^{\beta-1}(n-\alpha p - \beta q)!$, which will usually be smaller than $k!$, and you obtain a contradiction.
answered Sep 2, 2016 at 14:52
Jan-Christoph Schlage-PuchtaJan-Christoph Schlage-Puchta
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