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Suppose that $\lambda_1,\lambda_2,\lambda_3$ are partititions of $n$. When do there exist permutations $\sigma_1,\sigma_2,\sigma_3 \in S_n$ such that

(1) $\sigma_1\sigma_2\sigma_3$ is the identity;

(2) the $\sigma_i$ generate a transitive subgroup of $S_n$; and

(3) the cycle type of $\sigma_i$ is $\lambda_i$ for all $i$?

One can show that the total number of parts among the $\lambda_i$ must be an integer that is at most $n+2$ and that is congruent to $n$ modulo 2. Is this condition sufficient? If $\lambda_3 = (n)$ and the total number of parts of the $\lambda_i$ is exactly $n+2$, then the existence is known.

This problem is equivalent to the existence of an irreducible curve $X$ over the complex numbers and a nonconstant morphism $X \rightarrow \mathbb{P}^1$ whose branch locus is contained in $\{0,1,\infty\}$ (a Belyi map) and whose monodromy generators have cycle types $\lambda_1,\lambda_2,\lambda_3$.

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For $n=2,3,4$ the condition is sufficient by explicit checking, assuming I didn't make a calculation mistake. –  Will Sawin Apr 29 '13 at 3:42

1 Answer 1

up vote 2 down vote accepted

The answer is no. No such triples exist for $n = 4$, $\lambda_1 = (3, 1)$, $\lambda_2 = \lambda_3 = (2, 2)$. (Indeed, in $S_4$, the double transpositions and the identity form the Klein four subgroup.)

For other results on this problem, see for example http://arxiv.org/abs/math/0508434.

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