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Problem A5 from putname's competition 2009 asks to prove that there is no finite abelian group such that the product of order of its elements is equal to $2^{2009}$. Starting from this problem, for any group $G$ let $P(G)$ be the product of order of elements of $G$, i.e., $\prod_{x\in G}\mbox{O}(x)$. And consider the sets $$ A := \{ P(G) \; | \; G \mbox{ is a finite group } \} \mbox{ and } B := \{ P(G) \; | \; G \mbox{ is a finite abelian group } \}$$ (e.g., $2^{2009} \not \in B$ and clearly $3,5,7... \not \in A$ )

I have tried to describe $A,B$ in terms of their elements but it gets nowhere. So my goal is to find necessary or sufficient conditions for a nutural number to be an element of $A$ (or $B$). Even calculating $P(S_n)$ would be something! So any suggestion or ideal would be helpful.

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    $\begingroup$ I do not really understand the harsh reception of this question. But possibly it is only the way it is presented (phrasing it in the imperative like often done for excercises, and without motivation). $\endgroup$
    – user9072
    Commented Sep 10, 2013 at 12:15
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    $\begingroup$ Have you tried computing the first few $n$, looking for patterns, consulting the Online Encyclopedia of Integer Sequences? (Have you tried anything?) $\endgroup$
    – Gerry Myerson
    Commented Sep 10, 2013 at 12:45
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    $\begingroup$ But start with the smallest groups, those will give you the smallest values of $n$. Honestly, try something. $\endgroup$
    – Gerry Myerson
    Commented Sep 10, 2013 at 13:06
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    $\begingroup$ I guess this must mean "product of elements of all elements of $G$" rather than "product of orders of some elements of $G$", but the wording itself is completely unclear. $\endgroup$
    – Derek Holt
    Commented Sep 10, 2013 at 13:33
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    $\begingroup$ Just because it's HARD doesn't mean it's a good question! I am inclined to think it's not very natural. Outside of solving Putnam questions, would anyone care? $\endgroup$
    – Todd Trimble
    Commented Sep 10, 2013 at 14:50

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First of all, the Putnam exam problem is about 2-groups and I highly doubt that the generalization proposed above is in any way tractable. That having been said, for $S_n$, every permutation has a cycle type labeled by a partition $\lambda=(\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_n\geq0)$ of $n$. So

$$P(S_n)=\prod_{\lambda}\mathrm{lcm}(\lambda_1,\ldots,\lambda_n)^{|K_\lambda|}$$

where, $K_\lambda$ is the conjugacy class labelled by $\lambda$. Its order is $\frac{n!}{z_\lambda}$ where, for $\lambda=(1^{m_1},2^{m_2},\ldots)$, $z_\lambda=\prod_{i\geq 1} i^{m_i}(m_i!)$.

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    $\begingroup$ Shouldn't the $|K_\lambda|$ should be an exponent or did I just think through this answer too quickly? $\endgroup$
    – ARupinski
    Commented Sep 10, 2013 at 22:16
  • $\begingroup$ This is only the product of element orders. I think a closed form was expected. $\endgroup$
    – user37834
    Commented Sep 11, 2013 at 11:20
  • $\begingroup$ ARupinski: Right, $|K_\lambda|$ should be in the exponent. Silva: How much more explicit an answer do you think is possible in the $S_n$ case? $\endgroup$
    – David Hill
    Commented Sep 11, 2013 at 14:05

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