# Factorization of permutations.

Let $n,k$ be positive integers such that $3n=2k$ and $N = \lfloor \alpha n\rfloor$ for some constant $0<\alpha<1$. Let $S_{3n}$ denote the permutation group of order $3n$. Consider the following subsets of $S_{3n}$:

$\mathcal C_2$: set of $\sigma \in S_{3n}$ such that cycle decomposition of $\sigma$ consists of $k$ disjoint transpositions.

$\mathcal C_3$ : $\sigma \in S_{3n}$ such that cycle decomposition of $\sigma$ consists of $n$ disjoint $3$-cycles.

$S_{n,N}$ : $\sigma \in S_{3n}$ such that cycle decomposition of $\sigma$ consists of $N$ cycles, min cycle length at least $3$.

Now consider

$\mathcal S$ ={$(\sigma_3 , \sigma_2) \in \mathcal C_3 \times \mathcal C_2$ : $\sigma_3 \circ \sigma_2 \in S_{n,N}$}

Let $\Phi: \mathcal S \to S_{n,N}$ be the natural map such that $\Phi (\sigma_3,\sigma_2) = \sigma_3 \circ \sigma_2$

My questions are

1) Is $\Phi$ surjective? Or is the image most of $S_{n,N}$?

2) If not: will it be surjective if we vary $\alpha$ over a certain range?

3) Can we say anything about the asymptotic size of the fibers $\Phi^{-1}(\sigma)$ for $\sigma \in S_{n,N}$ for a typical $\sigma$?

Any references will be helpful. Thanks.

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Your question seems to be covered by the comments and answers to [mathoverflow.net/questions/62088/…. And unless I misunderstand what you are asking in (2), there is no way $\Phi$ can be surjective for an entire range of $\alpha$'s if it fails to be surjective for some $\alpha$ in the given range. –  ARupinski Jan 30 '13 at 4:38
@ARupinski: Thanks! I am looking into it. –  gmath Jan 30 '13 at 19:33