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iBac = materialul ULTRACOMPLET de pregătire pentru bac la mate. Dacă vrei poţi.
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[1]
Autor |
Mesaj |
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Scriu a in loc de alpha si s in loc de sigma.
Unde este problema de fapt?
Avem urmatorii pasi:
- Conditia de comutare cu a = (123)(456) este rescrisa folosind conditia de conjugare.
- Conjugarea cu s (sau cu inversul lui s...) aplicata pe o permutare scrisa ca produs de cicli disjuncti conduce la un ciclu de acelasi tip, s se aplica pe fiecare "simbol" in parte. Deci dam de ( s1, s2, s3 ) ( s4, s5, s6 ) .
(Am facut economie de paranteze la aplicare lui s pe cate un simbol.)
- Doua permutari scrise ca produs de cicli disjuncti sunt egale, daca si numai daca cicli de aceeasi lungime "se corespund" dupa o eventuala re-ordonare a lor si dupa o re-cilare in interiorul partilor care se corespund, data/fixata fiind o ordonare.
- La noi trebuie sa avem deci:
--- fie (123) = (s1 s2 s3) si (456) = (s4 s5 s6)
--- fie (456) = (s1 s2 s3) si (123) = (s4 s5 s6)
- Daca stim valoarea lui s1, atunci stim si s2 si s3:
--- s1 = 1 implica 1,2,3 "este" (ca tuplet) s1,s2,s3
--- s1 = 2 implica 2,3,1 "este" (ca tuplet) s1,s2,s3
--- s1 = 3 implica 3,1,2 "este" (ca tuplet) s1,s2,s3
--- s1 = 4 implica 4,5,6 "este" (ca tuplet) s1,s2,s3
--- s1 = 5 implica 5,6,4 "este" (ca tuplet) s1,s2,s3
--- s1 = 6 implica 6,4,5 "este" (ca tuplet) s1,s2,s3
- La fel si cu s4.
- Apoi autorul ia toate posibilitatile de alegere pentru valorile s1, s4 "de parti diferite" (nu ambele in 1,2,3 si nu ambele in 4,5,6, (de)ci una de o parte, cealalta pe cealalta parte).
Mai jos sunt alegerile si permutarea ce corespunde:
s1 s4 permutare
(ordinea este cea de mai sus - numai buna... Undeva am sters un ; )
14 e;
15 (4,5,6);
16 (4,6,5);
24 (1,2,3);
25 (1,2,3)(4,5,6);
26 (1,2,3)(4,6,5);
34 (1,3,2);
35 (1,3,2)(4,5,6);
36 (1,3,2)(4,6,5);
41 (1,4)(2,5)(3,6);
42 (1,4,2,5,3,6);
43 (1,4,3,6,2,5);
51 (1,5,2,6,3,4);
52 (1,5,3,4,2,6);
53 (1,5)(2,6)(3,4);
61 (1,6,3,5,2,4);
62 (1,6)(2,4)(3,5);
63 (1,6,2,4,3,5)
Recomand folosirea unui soft matematic...
Eu folosesc sage..
(Cand ajung acasa dau drumul...)
--- df (gauss)
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Cod sage...
sage: G = SymmetricGroup(6)
sage: g = PermutationGroupElement([(1,2,3),(4,5,6)])
sage: G
Symmetric group of order 6! as a permutation group
sage: G.order()
720
sage: g
(1,2,3)(4,5,6)
sage: g.order()
3
sage: for s in G.list():
....: if s*g == g*s:
....: print s
....:
()
(4,5,6)
(4,6,5)
(1,2,3)
(1,2,3)(4,5,6)
(1,2,3)(4,6,5)
(1,3,2)
(1,3,2)(4,5,6)
(1,3,2)(4,6,5)
(1,4)(2,5)(3,6)
(1,4,2,5,3,6)
(1,4,3,6,2,5)
(1,5,2,6,3,4)
(1,5,3,4,2,6)
(1,5)(2,6)(3,4)
(1,6,3,5,2,4)
(1,6)(2,4)(3,5)
(1,6,2,4,3,5)
Ceva mai structural... (mai departe)
sage: G.centralizer(g)
Subgroup of (Symmetric group of order 6! as a permutation group) generated by [(4,5,6), (1,2,3), (1,4)(2,5)(3,6)]
sage: H = G.centralizer(g)
sage: H.order()
18
sage: for s in H.list(): print s
()
(4,5,6)
(4,6,5)
(1,2,3)
(1,2,3)(4,5,6)
(1,2,3)(4,6,5)
(1,3,2)
(1,3,2)(4,5,6)
(1,3,2)(4,6,5)
(1,4)(2,5)(3,6)
(1,4,2,5,3,6)
(1,4,3,6,2,5)
(1,5,2,6,3,4)
(1,5,3,4,2,6)
(1,5)(2,6)(3,4)
(1,6,3,5,2,4)
(1,6)(2,4)(3,5)
(1,6,2,4,3,5)
Este H vreun grup "cunoscut" ?
Este de exemplu H izomorf cu grupul diedru cu tot atatea elemente?
sage: H.order()
18
sage: H.normal_subgroups()
[Subgroup of (Subgroup of (Symmetric group of order 6! as a permutation group) generated by [(4,5,6), (1,2,3), (1,4)(2,5)(3,6)]) generated by [()],
Subgroup of (Subgroup of (Symmetric group of order 6! as a permutation group) generated by [(4,5,6), (1,2,3), (1,4)(2,5)(3,6)]) generated by [(1,3,2)(4,5,6)],
Subgroup of (Subgroup of (Symmetric group of order 6! as a permutation group) generated by [(4,5,6), (1,2,3), (1,4)(2,5)(3,6)]) generated by [(1,3,2)(4,5,6), (1,4)(2,5)(3,6)],
Subgroup of (Subgroup of (Symmetric group of order 6! as a permutation group) generated by [(4,5,6), (1,2,3), (1,4)(2,5)(3,6)]) generated by [(1,3,2)(4,6,5)],
Subgroup of (Subgroup of (Symmetric group of order 6! as a permutation group) generated by [(4,5,6), (1,2,3), (1,4)(2,5)(3,6)]) generated by [(4,5,6), (1,2,3)],
Subgroup of (Subgroup of (Symmetric group of order 6! as a permutation group) generated by [(4,5,6), (1,2,3), (1,4)(2,5)(3,6)]) generated by [(4,5,6), (1,2,3), (1,4)(2,5)(3,6)]]
sage:
sage: D = DihedralGroup( 9 )
sage: D.order()
18
sage: D.normal_subgroups()
[Subgroup of (Dihedral group of order 18 as a permutation group) generated by [()],
Subgroup of (Dihedral group of order 18 as a permutation group) generated by [(1,4,7)(2,5,8)(3,6,9)],
Subgroup of (Dihedral group of order 18 as a permutation group) generated by [(1,2,3,4,5,6,7,8,9), (1,4,7)(2,5,8)(3,6,9)],
Subgroup of (Dihedral group of order 18 as a permutation group) generated by [(1,2,3,4,5,6,7,8,9), (1,9)(2,8)(3,7)(4,6)]]
sage: H.is_isomorphic( D )
False
Desigur ca nu... Grupul diedru are un element de ordin 9.
Elementele din H, centralizatorul lui g in G, grupul elementelor din problema, au ordinele urmatoare:
sage: for s in H: print "%s are ordinul %s" % ( s, s.order() )
() are ordinul 1
(4,5,6) are ordinul 3
(4,6,5) are ordinul 3
(1,2,3) are ordinul 3
(1,2,3)(4,5,6) are ordinul 3
(1,2,3)(4,6,5) are ordinul 3
(1,3,2) are ordinul 3
(1,3,2)(4,5,6) are ordinul 3
(1,3,2)(4,6,5) are ordinul 3
(1,4)(2,5)(3,6) are ordinul 2
(1,4,2,5,3,6) are ordinul 6
(1,4,3,6,2,5) are ordinul 6
(1,5,2,6,3,4) are ordinul 6
(1,5,3,4,2,6) are ordinul 6
(1,5)(2,6)(3,4) are ordinul 2
(1,6,3,5,2,4) are ordinul 6
(1,6)(2,4)(3,5) are ordinul 2
(1,6,2,4,3,5) are ordinul 6
Situatia pare a fi mai complicata.
Cu grupurile, pentru a le afla structura, putem sa le "facem bucatele", dar trebuie sa fim atenti. Trecerea la grup cât este posibila doar daca "luam modulo" un subgrup normal.
O sa ma leg de subgrupul normal al lui H care este generat de
(1,2,3)(4,5,6) . Iata structura si ce putem verifica repede:
sage: h = H( "(1,2,3)(4,5,6)" )
sage: h
(1,2,3)(4,5,6)
sage: K = H.subgroup( [h] )
sage: K
Subgroup of (Subgroup of (Symmetric group of order 6! as a permutation group) generated by [(4,5,6), (1,2,3), (1,4)(2,5)(3,6)]) generated by [(1,2,3)(4,5,6)]
sage: K.order()
3
sage: K.is_normal( H )
True
sage: L = H.quotient( K )
sage: L
Permutation Group with generators [(1,2)(3,6)(4,5), (1,3,5)(2,4,6)]
sage: L.order()
6
sage: L.is_isomorphic( SymmetricGroup(3) )
True
sage: L.is_isomorphic( DihedralGroup(3) )
True
Matematic scriem asa:
--- df (gauss)
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