6!\cdot 7!=10!6!\cdot 7!=10!. Is there a natural bijection between S_6\times S_7S_6\times S_7 and S_{10}S_{10}?

Aside from 1!\cdot n!=n! and (n!-1)!\cdot n! = (n!)!, the only nontrivial product of factorials known is 6!\cdot 7!=10!.

One might naturally associate these numbers with the permutations on 6, 7, and 10 objects, respectively, and hope that this result has some kind of connection to a sporadic relation between such permutations – numerical “coincidences” often have deep math behind them, like how 1^2+2^2+\ldots+24^2=70^2 can be viewed as an ingredient that makes the Leech lattice work.

The most natural thing to hope for would be a product structure on the groups S_6 and S_7 mapping to S_{10}, but as this MathOverflow thread shows, one cannot find disjoint copies of S_6 and S_7 living in S_{10}, so a product structure seems unlikely.

However, I’m holding out hope that some weaker kind of bijection can be found in a “natural” way. Obviously one can exhibit a bijection. For instance, identify the relative ordering of 1,2,\ldots 7 in a permutation of size 10, and then biject _{10}P_{3}=720 with S_6 in some way. But I’d like to know if there is a way to define such a bijection which arises naturally from the permutation structures on these sets, and makes it clear why the construction does not extend to other orders.

I tried doing something with orderings on polar axes of the dodecahedron (10!) and orderings on polar axes of the icosahedron (6!), in the hopes that the sporadic structure and symmetry of these Platonic solids would allow for interesting constructions that don’t generalize, but ran into issues with the dodecahedron (sequences of dodecahedral axes aren’t particularly nice objects) and the question of how to extract a permutation of length 7.

I’m curious if someone can either devise a natural bijection between these sets or link to previous work on this question.

Answer

This family of bijections (of sets) S_6\times S_7 \to S_{10} has already been suggested in comments and linked threads, but it is so pretty I wanted to spell it out:

There are 10 ways of partitioning the numbers 1,2,3,4,5,6 into two (unordered) pieces of equal size: P_1,P_2,\cdots,P_{10}. Thus we have a canonical embedding S_6\hookrightarrow S_{10}, coming from the induced action on the P_i.

Any distinct pair P_i,P_j will be related by a unique transposition. For example \{\{1,2,3\},\{4,5,6\}\} (denoted hereafter \left(\frac{123}{456}\right)) is related to \left(\frac{126}{453}\right) via the transposition (36).

There are two types of ordered (distinct) triples P_i, P_j,P_k:

  1. They may be related pairwise via transpositions (ab),(cd),(ef) with a,b,c,d,e,f distinct and each of \{a,b\}, \{c,d\},\{e,f\} not on the same side of any of P_i, P_j,P_k:
    \left(\frac{ace}{bdf}\right), \left(\frac{bce}{adf}\right), \left(\frac{ade}{bcf}\right).

    Here, there are 10 choices for P_i, 9 choices for P_j and 4 choices for P_k, giving 360 triples in total.

  2. They may be related pairwise via transpositions (ab),(bc),(ca) with a,b,c distinct:

    \left(\frac{ace}{bdf}\right), \left(\frac{bce}{adf}\right), \left(\frac{abe}{cdf}\right).

    Again, there are 10 choices for P_i, 9 choices for P_j and 4 choices for P_k, giving 360 triples in total.

An element of the stabiliser (in S_6) of a type 1 ordered triple (written as above) must preserve the pairs \{a,b\}, \{c,d\},\{e,f\}. Further if it swaps any of these pairs it must swap all of them, so the only non-trivial element of the stabiliser is an odd permutation: (ab)(cd)(ef).

An element of the stabiliser (in S_6) of a type 2 ordered triple (written as above) must preserve the sets \{d,f\}, \{e\},\{a,c,b\}. Further it must fix each of a,b,c. Thus the only non-trivial element of the stabiliser is an odd permutation: (df).

As |A_6|=360, in particular this means there is a unique element of A_6 taking the ordered triple P_1,P_2,P_3 to a specified ordered triple P_i,P_j,P_k of the same type as P_1,P_2,P_3.

Fix t\in S_{10} a permutation taking P_1,P_2,P_3 to an ordered triple of the other type. Then there is a unique element in A_6 which composed with t takes the ordered triple P_1,P_2,P_3 to a specified ordered triple P_i,P_j,P_k of the other type to P_1,P_2,P_3.

Let S_7 denote the group of permutations of P_4,P_5,\cdots,P_{10}. Then any permutation in S_{10} may be written uniquely as an element of S_7 followed by an element of (A_6\sqcup tA_6), where the latter is determined by where P_1,P_2,P_3 are mapped to.

Thus we have established a bijection of sets S_{10}\to (A_6\sqcup tA_6)\times S_7.
Once we fix an odd permutation t’\in S_6, we may identify the sets (A_6\sqcup t’A_6)\to S_6.
Composing we get: S_{10}\to (A_6\sqcup tA_6)\times S_7\to (A_6\sqcup t’A_6)\times S_7\to S_6\times S_7.

That is for any choice of the permutations t,t’ we have the required bijection of sets.

Attribution
Source : Link , Question Author : RavenclawPrefect , Answer Author : tkf

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