65
$\begingroup$

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.

$\endgroup$
  • 10
    $\begingroup$ The linked MathOverflow question mentions that there exists a knit product of $S_7$ and the Mathieu group $M_{10}$ giving $S_{10}$. There is an isomorphism $M_{10} \simeq A_6.C_2$, where $A_6$ is the alternating group on $6$ elements; the construction seems to be related to the exceptional outer automorphism of $S_6$. For example, Section 6.2 here describes the construction, starting from the $10$ partitions of a set of $6$ elements into two parts of three. $\endgroup$ – pregunton Nov 11 at 18:28
  • 2
    $\begingroup$ (continued) Perhaps some geometric description can be found, as you suggest, by identifying that $6$-element set with the $6$ polar axes of an icosahedron, similarly to the discussion here. $\endgroup$ – pregunton Nov 11 at 18:28
  • 4
    $\begingroup$ Here's a thought: An element $\pi \in S_{10}$ is completely specified by: 1. an element of $\sigma \in S_7$ corresponding to the relative order of $\pi(1),\dots,\pi(7)$, 2. a subset $R \subset [10] = \{1,\dots,10\}$ of size $7$, namely $\{\pi(1),\dots,\pi(7)\}$, 3. a bijection from $[10]\setminus R$ to $\{8,9,10\}$. Equivalently, an element $\tau \in S_3$. With that, we have a bijection between $S_{10}$ and $S_7 \times \binom{[10]}{3} \times S_3$. So, it would suffice to find a bijection between $S_6$ and $\binom{[10]}{3} \times S_3$. $\endgroup$ – Ben Grossmann Nov 11 at 18:35
  • 4
    $\begingroup$ Another possibly relevant numerical coincidence: $6$ is the third triangular number, and $10$ is the third tetrahedral number and also the fourth triangular number. $\endgroup$ – Ben Grossmann Nov 11 at 18:38
  • 1
    $\begingroup$ This doesn't feel particularly natural to me? An approach based on decomposing prime factors seems mostly like a verification of the fact that $6!\cdot 7!=10!$ the usual way, whereas I'm imagining something where a natural sporadic symmetry shows that the product has to hold, without actually checking that the counts of 2s and 3s and 5s and 7s add up right. It doesn't seem like it yields insight into why the bijection works for $6,7,$ and $10$ but not some other triple of numbers. $\endgroup$ – RavenclawPrefect Nov 21 at 0:54
4
+50
$\begingroup$

It may be connected with, of all things, the $3-4-5$ right triangle! This triangle and its multiples stand out as having the sides in arithmetic progression. Such an arithmetic progression leads to factorial expressions when the sides are multiplied together.

As a preliminary step, consider a relatively unheralded property of right triangles: the diameter of the incircle plus the hypotenuse equals the sum of the other two sides. Suppose that the legs are $a$ and $b$, and the hypoteneuse is $c$ where $c^2=a^2+b^2$. The diameter of the incircle is then $2ab/(a+b+c)$ while the Pythagorean relation implies $$(a+b+c)(a+b-c)=(a^2+2ab+b^2)-(a^2+b^2)=2ab$$ Thereby the diameter of the incircle reduces to $a+b-c$. Should there be a right triangle whose sides are in arithmetic progression, then, the diameter of the incircle will join this progression, making it longer and thus perhaps generating a bigger factorial upon multiplication.

In this question it is shown that the product of the sides of any triangle is half the product of the diameter of the circumcircle (circumdiameter), the diameter of the incircle (indiameter), and the perimeter. Let us see where that leads if we apply it to a right triangle having sides $3,4,5$. Multiplying the sides together then gives

$3×4×5=\text{circumdiameter}×\text{indiameter}×\text{perimeter}/2$

We double the sides of the triangle to clear the fraction on the right side:

$6×8×10=\text{circumdiameter}×\text{indiameter}×\text{perimeter}×4$

The circumdiameter is the hypoteneuse of the $3-4-5$ triangle, thus $5$ -- which is in the aforementioned arithmetic progression. The indiameter is $2$ from the above lemma, which precedes $3,4,5$ in the arithmetic progression. And the perimeter of the triangle is three times the longer leg, again due to the arithmetic progression, thus $4×3$. Substituting these results into the above product equality then gives

$6×8×10=5×2×(3×4)×4=5!×4$

And there is our factorial. To make it cleaner we should multiply by $3/2$, absorbing the dangling factor $4$ into the factorial. We then get three different three-term products on the left side, depending on which of the factors $6,8,10$ we increment:

$\color{blue}{8×9×10}=6×10×12=6×8×15=5×2×(3×4)×6=6!$

And from the three-term product shown in blue, we have

$6!=10!/7!$

Why is this uniquely chosen? We see that the sides of a right triangle being in arithmetic progression lead to the factorial on the right in two ways, by making the perimeter a simple multiple of one leg and by incorporating the circum diameter into the arithmetic progression. Only the $3-4-5$ right triangle has these properties, and it leads specifically to $6!$ also being a factorial ratio.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Interesting derivation of the fact that $6!\cdot 7!=10!$ - I don’t see how this gives rise to a natural bijection, though? $\endgroup$ – RavenclawPrefect Nov 21 at 18:49
  • $\begingroup$ It does not give a bi-jection because I use algebra and geometry instead of group theory. Instead, the arithmetic progression of the incircle diameter and the sides, unique to one triangle, gives a coalescence of factors that lead to a factorial product relation. When I try the same methodology with other right triangles where there isn't this arithmetic progression, the factors are scattered and can't elegantly be combined. $\endgroup$ – Oscar Lanzi Nov 21 at 20:27
  • 2
    $\begingroup$ Although I appreciate the novel view on the product given in the question, I'm not planning to award this answer the bounty, as I'm really seeking a natural bijection here. (By default I believe this question will receive half the bounty, and I wished to clarify that this default action isn't due to my forgetting about the bounty.) $\endgroup$ – RavenclawPrefect Nov 28 at 7:33
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Interesting though my geometrically based approach may be, the OP was looking for one based on group theory, so if this takes the bounty points that's fair. +1. $\endgroup$ – Oscar Lanzi Nov 29 at 14:55
  • 1
    $\begingroup$ @OscarLanzi Thankyou, that is very generous of you to say. I enjoyed reading your answer, and learnt a lot about triangles. $\endgroup$ – tkf Nov 29 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.