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Suppose $n$ is a natural number and $\lambda$ is an unordered integer partition of $n$ such that $\lambda$ has $a_{\lambda,j}$ parts of size $j$ for each $j$... Let $c_\lambda$ be the conjugacy class in the symmetric group of degree $n$ comprising the elements whose cycle type is $\lambda$... Then: $$ \! |c_\lambda| = \frac{n!}{\prod_j (j)^{a_{\lambda,j}}(a_{\lambda,j}!)} $$ (see here for more information: Conjugacy class size formula in symmetric group)

Let $\Lambda$ be the set of all partitions $\lambda$ and $M\subset\Lambda $. For $n\ge5$ there seem to be more than one solution $M$ for the following: $$ \sum_{\mu\in M} |c_\mu|=\frac{n!}2 \tag{1} $$

  • There is at least one solution, since the splitting into odd and even permutations divides the set of all permutations properly.

  • For $n=5$ the even/odd splitting looks like $$ \begin{eqnarray} |c_{(1,1,1,1,1)}|+|c_{(3,1,1)}|+|c_{(2,2,1)}|+|c_{(5)}|&=&1+20+15+24\\ |c_{(2,1,1,1)}|+|c_{(4,1)}|+|c_{(3,2)}|&=&10+30+20, \end{eqnarray} $$ but there are at least $3$ more solutions:

    1. Since $|c_{3,2}|=\frac{5!}{(3^1\cdot 1!)(2^1\cdot 1!)}=|c_{3,1,1}|=\frac{5!}{(3^1\cdot 1!)(1^2\dot 2!)}$ you can interchange the odd conjugacy class of the partition $3+2$ with the even one for the partition $3+1+1$.

    2. and 3. $|c_{(1,1,1,1,1)}|+|c_{(2,2,1)}|+|c_{(5)}|=1+15+24=40$, so we may again add $|c_{(3,2)}|$ or $|c_{(3,1,1)}|$

My question:

Is it possible to give an asymptotic? How many solutions $M$ exist for $(1)$ in general?

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Ok, it was easier than I thought to find a concise way to calculate the number of solutions:

We've got to calculate $$ \prod_{\lambda\in\Lambda} \left(1+x^{|c_\lambda|}\right) $$

and then find the coefficient of $\displaystyle x^\frac{n!}{2}$. Some results:

  • $n=5$ (supported by W|A):

$$(1+x)(1+x^{10})(1+x^{20})^2 (1+x^{15})(1+x^{30})(1+x^{24})=x^{120}+x^{119}+x^{110}+x^{109}+x^{105}+x^{104}+2 x^{100}+2 x^{99}+x^{96}+2 x^{95}+x^{94}+3 x^{90}+3 x^{89}+x^{86}+3 x^{85}+2 x^{84}+x^{81}+3 x^{80}+2 x^{79}+2 x^{76}+5 x^{75}+3 x^{74}+x^{71}+4 x^{70}+3 x^{69}+3 x^{66}+5 x^{65}+2 x^{64}+2 x^{61}+\color{red}{4 x^{60}}+2 x^{59}+2 x^{56}+5 x^{55}+3 x^{54}+3 x^{51}+4 x^{50}+x^{49}+3 x^{46}+5 x^{45}+2 x^{44}+2 x^{41}+3 x^{40}+x^{39}+2 x^{36}+3 x^{35}+x^{34}+3 x^{31}+3 x^{30}+x^{26}+2 x^{25}+x^{24}+2 x^{21}+2 x^{20}+x^{16}+x^{15}+x^{11}+x^{10}+x+1$$

  • $n=6$: $$ (1+x^{1})(1+x^{15})(1+x^{40})^2(1+x^{90})^2(1+x^{45})(1+x^{144})(1+x^{120})^2(1+x^{15})\\=\dots+ \color{red}{12 x^{360}}+\dots $$

  • $n=7$: $(1+x) \dots (1+x^{840}) = \dots+ \color{red}{58 x^{2520}}+\dots$

  • $n=8$: $(1+x) \dots (1+x^{5760}) = \dots+ \color{red}{1016 x^{20160}}+\dots$

So summarizing this we get: $0,2,2,2,4,12,58,1016$ (thanks to Jack) for the first eight symmetric groups. The question about the asymptotics remains open...

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Some of your earlier terms are wrong, I think. 0, 2, 2, 2, 4, 12, 58, 1016, 53940, ... If you require the identity, then you get 0, 1, 1, 1, 2, 6, 29, 508, 26970, ... (the relationship is kind of neat). –  Jack Schmidt Jun 5 '13 at 16:57
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next term is 12653364 (or 6326682 if you require the identity; eerie). The consecutive ratios are ∞, 1, 1, 2, 3, 4.8, 17.5, 53.1, 234.6, so that looks super-exponential to me. At any rate, the lesson that numerical coincidences might be helpful in A5, but not in general is pretty clear. :-) –  Jack Schmidt Jun 5 '13 at 19:01
    
@jack, I see that my approach isn't as good as I thought. Gotta think about that: Isn't it the same problem as here: Making Change for a Dollar (and other number partitioning problems)? Ok I didn't use the reciprocal. Do I have to, to solve this problem? But I don't get what you mean by requiring identity... –  draks ... Jun 5 '13 at 19:43
    
Requiring identity: You claimed that there was only one way to find a union of conjugacy classes of size 1 in the symmetric group S2, but obviously there are two ways {()} and {(1,2)} both work. I tried to guess why your count was wrong. I guessed that it was because you were trying to write the alternating group as a union of conjugacy classes, and obviously every subgroup has to the identity, so maybe you wanted that conjugacy class always used. Note that $(1+x)(1+x) = 1 +\color{red}{2x^1} + x^2$, so the polynomial method gives 2 as well. –  Jack Schmidt Jun 5 '13 at 20:01
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"Super exponential" in the complexity class sense. Faster than exponential, $f:\mathbb{N} \to \mathbb{R}$ is super exponential if for every $C>0, D>1$ there is some $N$ such that $f(n) > C \cdot D^n$ for all $n \geq N$. Note that $(C\cdot D^{n+1})/(C\cdot D^n) = D$, so the consecutive ratios give an estimate for $D$ (namely, infinity). –  Jack Schmidt Sep 11 '13 at 14:40
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