Een oplossing met MuPAD light 2.5

Eerst voeren we de betreffende permutatiegroep G in, voortgebracht door de 4-cykels a=(1234) en b=(3456).

G := Dom::PermutationGroup(6,[[[1,2,3,4]],[[3,4,5,6]]])
Dom::PermutationGroup(6, [[[1, 2, 3, 4]], [[3, 4, 5, 6]]])

Vervolgens bepalen we zijn orde ("size", "order" is gereserveerd voor de orde van een individueel element)

G::size;
120

Nu zijn we al bijna klaar, want de hoogste macht van 3 die in 120 opgaat is 3. Alle Sylow 3-subgroepen van G zijn dus 3-cyclisch (en omgekeerd). Daar bij voorbeeld s = ba =(124)(356), is S = {1,s,s-1} zo'n Sylow subgroep. Maar algemeen geldt in een (eindige) groep dat (voor een vast priemgetal p) alle Sylow p-subgroepen onderling geconjugeerd zijn; in het bijzonder zijn alle andere elementen van de orde 3 met s of s-1 = (142)(365) geconjugeerd en hebben dus zeker dezelfde cykelstructuur.

Natuurlijk kunnen we ook direct uit alle elementen van G:

L := G::allElements
[[6, 5, 4, 3, 2, 1], [6, 5, 2, 1, 4, 3], [2, 1, 4, 3, 6, 5], [4, 3, 6, 5, 2, 1], [3, 2, 1, 4, 6, 5], [6, 5, 3, 2, 1, 4], [6, 5, 1, 4, 3, 2], [2, 1, 3, 6, 5, 4], [1, 4, 3, 2, 6, 5], [4, 3, 1, 6, 5, 2], [3, 6, 5, 4, 2, 1], [1, 6, 5, 2, 4, 3], [6, 3, 5, 1, 2, 4], [2, 1, 5, 4, 3, 6], [3, 2, 6, 5, 1, 4], [5, 4, 2, 1, 3, 6], [6, 4, 1, 5, 2, 3], [5, 4, 3, 6, 2, 1], [1, 4, 2, 6, 5, 3], [3, 2, 5, 1, 4, 6], [4, 3, 2, 1, 6, 5], [3, 2, 4, 6, 5, 1], [2, 6, 5, 3, 1, 4], [6, 4, 5, 2, 3, 1], [4, 6, 5, 1, 3, 2], [6, 2, 1, 3, 5, 4], [6, 2, 5, 4, 1, 3], [1, 4, 5, 3, 2, 6], [1, 6, 3, 5, 2, 4], [6, 4, 3, 1, 5, 2], [3, 6, 1, 5, 4, 2], [2, 1, 6, 5, 4, 3], [6, 1, 3, 4, 2, 5], [5, 3, 1, 4, 2, 6], [6, 3, 4, 5, 1, 2], [4, 2, 1, 5, 3, 6], [5, 4, 6, 2, 1, 3], [5, 3, 2, 6, 1, 4], [1, 5, 4, 2, 3, 6], [5, 6, 4, 1, 2, 3], [1, 3, 6, 2, 5, 4], [3, 5, 1, 6, 2, 4], [5, 4, 1, 3, 6, 2], [6, 1, 4, 2, 5, 3], [1, 3, 2, 5, 4, 6], [6, 1, 5, 3, 4, 2], [4, 3, 5, 2, 1, 6], [4, 6, 2, 5, 1, 3], [6, 3, 2, 4, 5, 1], [6, 2, 4, 1, 3, 5], [2, 6, 4, 5, 3, 1], [5, 1, 6, 3, 2, 4], [1, 4, 6, 5, 3, 2], [3, 6, 2, 1, 5, 4], [6, 4, 2, 3, 1, 5], [5, 2, 4, 3, 1, 6], [6, 2, 3, 5, 4, 1], [3, 1, 4, 5, 2, 6], [1, 6, 4, 3, 5, 2], [5, 3, 6, 1, 4, 2], [5, 2, 1, 6, 4, 3], [4, 6, 1, 3, 2, 5], [4, 5, 3, 1, 2, 6], [5, 6, 3, 4, 1, 2], [4, 2, 6, 1, 5, 3], [2, 5, 4, 6, 1, 3], [5, 3, 4, 2, 6, 1], [1, 5, 6, 4, 2, 3], [1, 3, 4, 6, 2, 5], [3, 5, 4, 1, 6, 2], [2, 6, 1, 4, 5, 3], [6, 3, 1, 2, 4, 5], [5, 1, 3, 2, 4, 6], [6, 1, 2, 5, 3, 4], [2, 4, 3, 5, 1, 6], [4, 6, 3, 2, 5, 1], [1, 6, 2, 4, 3, 5], [5, 2, 6, 4, 3, 1], [5, 1, 4, 6, 3, 2], [3, 6, 4, 2, 1, 5], [3, 5, 2, 4, 1, 6], [5, 6, 2, 3, 4, 1], [3, 1, 6, 4, 5, 2], [1, 5, 3, 6, 4, 2], [5, 2, 3, 1, 6, 4], [3, 4, 6, 1, 2, 5], [4, 5, 6, 3, 1, 2], [4, 2, 3, 6, 1, 5], [2, 5, 3, 4, 6, 1], [4, 1, 5, 6, 2, 3], [1, 3, 5, 4, 6, 2], [2, 6, 3, 1, 4, 5], [2, 5, 1, 3, 4, 6], [5, 6, 1, 2, 3, 4], [2, 4, 6, 3, 5, 1], [4, 1, 6, 2, 3, 5], [4, 5, 2, 6, 3, 1], [5, 1, 2, 4, 6, 3], [2, 3, 6, 4, 1, 5], [3, 5, 6, 2, 4, 1], [3, 1, 2, 6, 4, 5], [1, 5, 2, 3, 6, 4], [3, 4, 5, 6, 1, 2], [4, 2, 5, 3, 6, 1], [4, 1, 3, 5, 6, 2], [1, 2, 3, 4, 5, 6], [1, 2, 6, 3, 4, 5], [2, 5, 6, 1, 3, 4], [2, 4, 1, 6, 3, 5], [4, 5, 1, 2, 6, 3], [2, 3, 5, 6, 4, 1], [3, 1, 5, 2, 6, 4], [3, 4, 2, 5, 6, 1], [4, 1, 2, 3, 5, 6], [1, 2, 5, 6, 3, 4], [2, 4, 5, 1, 6, 3], [2, 3, 1, 5, 6, 4], [3, 4, 1, 2, 5, 6], [1, 2, 4, 5, 6, 3], [2, 3, 4, 1, 5, 6]]

de elementen van de orde 3 selecteren.

ord3:=select(L,G::order=3)
[[6, 3, 5, 1, 2, 4], [6, 4, 1, 5, 2, 3], [6, 1, 5, 3, 4, 2], [4, 6, 2, 5, 1, 3], [2, 6, 4, 5, 3, 1], [5, 1, 6, 3, 2, 4], [6, 4, 2, 3, 1, 5], [5, 3, 6, 1, 4, 2], [4, 6, 1, 3, 2, 5], [2, 5, 4, 6, 1, 3], [5, 3, 4, 2, 6, 1], [3, 5, 4, 1, 6, 2], [5, 6, 1, 2, 3, 4], [4, 1, 6, 2, 3, 5], [4, 5, 2, 6, 3, 1], [3, 5, 6, 2, 4, 1], [3, 1, 2, 6, 4, 5], [3, 4, 5, 6, 1, 2], [2, 4, 5, 1, 6, 3], [2, 3, 1, 5, 6, 4]]

en die vervolgens in disjuncte cykels opsplitsen.

Order3:=map(ord3,G::cycles)
[[[1, 6, 4], [2, 3, 5]], [[1, 6, 3], [2, 4, 5]], [[1, 6, 2], [3, 5, 4]], [[1, 4, 5], [2, 6, 3]], [[1, 2, 6], [3, 4, 5]], [[1, 5, 2], [3, 6, 4]], [[1, 6, 5], [2, 4, 3]], [[1, 5, 4], [2, 3, 6]], [[1, 4, 3], [2, 6, 5]], [[1, 2, 5], [3, 4, 6]], [[1, 5, 6], [2, 3, 4]], [[1, 3, 4], [2, 5, 6]], [[1, 5, 3], [2, 6, 4]], [[1, 4, 2], [3, 6, 5]], [[1, 4, 6], [2, 5, 3]], [[1, 3, 6], [2, 5, 4]], [[1, 3, 2], [4, 6, 5]], [[1, 3, 5], [2, 4, 6]], [[1, 2, 4], [3, 5, 6]], [[1, 2, 3], [4, 5, 6]]]

We zien, expliciet, dat inderdaad uitsluitend producten van 2 disjuncte 3-cykels voorkomen. In het bijzonder herkennen we s=(124)(356) en zijn inverse.

Hoe komen we aan A,B,...,E in de gepubliceerde oplossing?

De S6, d.w.z de volledige permutatiegroep van {1,2,3,4,5,6}, heeft net zo veel 2-cykels als producten van 2k+1 disjuncte 2-cykels, namelijk 15 (met k=1). (Deze combinatorische "toevalligheid" heeft tot gevolg dat de S6 als enige(!) onder de Sn uitwendige automorfismes heeft. We laten het verder bij dit topje van de ijsberg.) Van deze 15 producten van 3 disjuncte cykels liggen er 10 in G en de overige 5 corresponderen precies met A,B,...,E.

oddord2:=select(L,G::order=2 and G::sign=-1)
[[6, 5, 4, 3, 2, 1], [2, 1, 4, 3, 6, 5], [4, 3, 2, 1, 6, 5], [6, 4, 5, 2, 3, 1], [4, 6, 5, 1, 3, 2], [3, 6, 1, 5, 4, 2], [2, 1, 6, 5, 4, 3], [5, 4, 6, 2, 1, 3], [5, 3, 2, 6, 1, 4], [3, 5, 1, 6, 2, 4]]

Oddorder2:=map(oddord2,G::cycles)
[[[1, 6], [2, 5], [3, 4]], [[1, 2], [3, 4], [5, 6]], [[1, 4], [2, 3], [5, 6]], [[1, 6], [2, 4], [3, 5]], [[1, 4], [2, 6], [3, 5]], [[1, 3], [2, 6], [4, 5]], [[1, 2], [3, 6], [4, 5]], [[1, 5], [2, 4], [3, 6]], [[1, 5], [2, 3], [4, 6]], [[1, 3], [2, 5], [4, 6]]]

S6:=Dom::SymmetricGroup(6)
Dom::SymmetricGroup(6)

t6:=select(map(select(S6::allElements,S6::order=2),S6::cycles),nops=3)
[[[1, 2], [3, 4], [5, 6]], [[1, 3], [2, 4], [5, 6]], [[1, 4], [2, 3], [5, 6]], [[1, 2], [3, 5], [4, 6]], [[1, 3], [2, 5], [4, 6]], [[1, 5], [2, 3], [4, 6]], [[1, 2], [3, 6], [4, 5]], [[1, 4], [2, 5], [3, 6]], [[1, 5], [2, 4], [3, 6]], [[1, 3], [2, 6], [4, 5]], [[1, 4], [2, 6], [3, 5]], [[1, 5], [2, 6], [3, 4]], [[1, 6], [2, 3], [4, 5]], [[1, 6], [2, 4], [3, 5]], [[1, 6], [2, 5], [3, 4]]]

AEDCB:=coerce(t6,DOM_SET) minus coerce(Oddorder2,DOM_SET)
{[[1, 2], [3, 5], [4, 6]], [[1, 3], [2, 4], [5, 6]], [[1, 4], [2, 5], [3, 6]], [[1, 5], [2, 6], [3, 4]], [[1, 6], [2, 3], [4, 5]]}

We herkennen A, E, D, C en B, in deze volgorde.

In feite is G (met orde 120 = 5!) isomorf met Sym({A,B,C,D,E}), de volledige permutatiegroep van {A,B,C,D,E}, ofwel met S5. Anders gezegd het door de werking gegeven homomorfisme f van G in Sym({A,B,C,D,E}) voortgebracht door f((1234))=(ABCD) en f((3456))=(BEDC) is bijectief.

Bewijs: Daar beide groepen dezelfde orde (namelijk 120) hebben is het voldoende de surjectiviteit van f aan te tonen. Welnu, omdat naast (ABCD) ook f(s)(= f((3456))f((1234))) = (AED) en (ABCD)(BEDC)-1 = (ABDEC) in de beeldgroep Im(f) van f liggen, is de orde van Im(f) een veelvoud van 4,3, en 5, dus (hier)gelijk aan 60 of 120. De eerste mogelijkheid vervalt, omdat Im(f) ook oneven permutaties bevat.