Simplicity of A5 Brute Force Method!

Transcription

1 Simplicity of A5 Brute Force Method! Step 1: List all the elements of A5. i (12)(12) (135) (13)(35) (12354) (12)(23)(35)(54) (12)(34) (12)(34) (145) (14)(45) (13425) (13)(34)(42)(25) (12)(35) (12)(35) (234) (23)(34) (15243) (15)(52)(24)(43) (12)(45) (12)(45) (235) (23)(35) (14532) (14)(45)(53)(32) (13)(24) (13)(24) (245) (24)(45) (12534) (12)(25)(53)(34) (13)(25) (13)(25) (345) (34)(45) (15423) (15)(54)(42)(23) (13)(45) (13)(45) (132) (13)(23) (13245) (13)(32)(24)(45) (14)(23) (14)(23) (142) (14)(24) (14352) (14)(43)(35)(52) (14)(25) (14)(25) (152) (15)(52) (12543) (12)(25)(54)(43) (14)(35) (14)(35) (143) (14)(34) (15324) (15)(53)(32)(24) (15)(23) (15)(23) (153) (15)(35) (14235) (14)(42)(23)(35) (15)(24) (15)(24) (154) (15)(45) (13452) (13)(34)(45)(52) (15)(34) (15)(34) (243) (24)(34) (12435) (12)(24)(43)(35) (23)(45) (23)(45) (253) (25)(35) (14523) (14)(45)(52)(23) (24)(35) (24)(35) (254) (25)(45) (13254) (13)(32)(25)(54) (25)(34) (25)(34) (354) (35)(45) (15342) (15)(53)(34)(42) (123) (12)(23) (12345) (12)(23)(34)(45) (12453) (12)(24)(45)(53) (124) (12)(24) (13524) (13)(35)(52)(24) (14325) (14)(43)(32)(25) (125) (12)(25) (14253) (14)(42)(25)(53) (15234) (15)(52)(23)(34) (134) (13)(34) (15432) (15)(54)(43)(32) (13542) (13)(35)(54)(42) Now, let s list all the subgroups! 1. {i} That s all the ones of order 1! 2. (12)(34) 3. (12)(35) 4. (12)(45) 5. (13)(24) 6. (13)(25) 7. (13)(45) 8. (14)(23) 9. (14)(25) 10. (14)(35) 11. (15)(23) 12. (15)(24) 13. (15)(34) 14. (23)(45) 15. (24)(35) 16. (25)(34) That s all the ones of order (123) 18. (124) 19. (125) 20. (134) 21. (135) 22. (145) 23. (234) 24. (235) 25. (245)

2 26. (345) That s all the ones of order (12345) 28. (12354) 29. (12534) 30. (12543) 31. (12435) 32. (12454) That s all the ones of order (12)(34),(13)(24) 34. (12)(35),(13)(25) 35. (12)(45),(14)(25) 36. (13)(45),(14)(35) 37. (23)(45),(24)(35) That s all the ones of order (34)(12),(35)(12) = (34)(12),(345) 39. (24)(13),(25)(13) = (24)(13),(245) 40. (23)(14),(25)(14) = (23)(14),(235) 41. (23)(15),(24)(15) = (23)(15),(234) 42. (14)(23),(15)(23) = (14)(23),(145) 43. (13)(24),(15)(24) = (13)(24),(135) 44. (13)(25),(14)(25) = (13)(25),(134) 45. (12)(34),(15)(34) = (13)(25),(125) 46. (12)(35),(14)(35) = (12)(35),(124) 47. (12)(45),(13)(45) = (12)(35),(123) That s all the ones of order 6, these are all isomorphic to D6. Note A5 can t have a commuting pair consisting of an order- 3 and order- 2 element. So, there are no subgroups isomorphic to Z (13)(45),(12345) 49. (13)(45),(12354) 50. (15)(34),(12534) 51. (15)(34),(12543) 52. (14)(35),(12435) 53. (14)(35),(12453) That s all the ones of order 10, these are all isomorphic to D10. Using similar logic as above, we also can t have a subgroup isomorphic to Z (12)(34),(123) 55. (12)(35),(123) 56. (12)(45),(124) 57. (13)(45),(134) 58. (23)(45),(234) That s all the ones of order 12. We know we can only have this one kind (ones isomorphic to A4) because it is the only kind of group of order 12 that doesn t have any elements of order A4 We can t have any subgroups of order 15 because we can t a commuting pair consisting of a 3- cycle and a 5- cycle. This means we can t have a subgroup of order 30 because all groups of order 30 contain subgroups of order 15. It s harder to explain why we can t have a subgroup of order 20. The reason is that none of the subgroups of order 4 are in the normalizer of any of the subgroups of order 5 (so we can t form a semi- direct product of a subgroup of order 4 and a subgroup of order 5 the order 5 subgroup would have to be normal in this big subgroup, which would mean the elements of the order 4 subgroup would have to be in its normalizer). I ll show that this is true when I do the conjugation checks on the order 5 subgroups. So, that s all the subgroups! Let s see if any are normal.

3 First, none of the order- 3 subgroups can be normal because you can always take a 3- cycle to a 3- cycle from a different subgroup of order 3 by conjugation, like so: (abd)(abc)(adb) = (bdc) Second, none of the order- 2 subgroups can be normal because you can always take a pair of disjoint transpositions to a different (non- trivial) pair of disjoint transpositions under conjugations, like so: (abe)(ab)(cd)(aeb) = (be)(cd) So, that s 26 of them eliminated! Now I want to show two things about the order 5 subgroups. 1) I want so show that none of them are normal. 2) I want to show that none of the order 4 subgroups are in the normalizer if an order 5 subgroup. 1) will follow from 2), so I ll just show two. Consider (12345): (12)(34)(12345) (12)(34) = (14352) which is not in (12345) So, (12)(34),(13)(24) is not in the normalizer of (12345). (12)(35)(12345) (12)(35) = (15432) which IS in (12345) BUT (13)(25)(12345) (13)(25) = (14235) which is not in (12345) So, (12)(35),(13)(25) is not in the normalizer of (12345). (12)(45)(12345) (12)(45) = (13542) which is not in (12345) So, (12)(45),(14)(25) is not in the normalizer of (12345). (13)(45)(12345) (13)(45) = (15432) which IS in (12345) BUT (14)(35)(12345) (14)(35) = (13425) which is not in (12345) So, (13)(45),(14)(35) is not in the normalizer of (12345). (23)(45)(12345) (23)(45) = (13254) which is not in (12345) So, (23)(45),(24)(35) is not in the normalizer of (12345). So, none of the subgroups of order 4 are in the normalizer of (12345). From above, we can already see that (12)(34),(13)(24) is not in the normalizer of (14352) = (12534). Let s continue working with (12534).

4 (12)(35)(12534) (12)(35) = (13542) which is not in (12534) So, (12)(35),(13)(25) is not in the normalizer of (12534). (14)(25)(12534) (14)(25) = (14523) which is not in (12534) So, (12)(45),(14)(25) is not in the normalizer of (12534). (13)(45)(12534) (13)(45) = (15324) which is not in (12534) So, (13)(45),(14)(35) is not in the normalizer of (12534). (23)(45)(12534) (23)(45) = (13425) which is not in (12534) So, (24)(35),(23)(45) is not in the normalizer of (12534). So, none of the subgroups of order 4 are in the normalizer of (12534). From above, we saw that (12)(35),(13)(25) is not in the normalizer of (14235) = (12543). We also just saw that (13)(45),(14)(35) is not in the normalizer of (15324) = (12543). Let s keep working with (12543). (12)(34)(12543) (12)(34) = (15342) which is not in (12543) So, (12)(34),(13)(24) is not in the normalizer of (12543). (12)(45)(12543) (12)(45) = (14532) which is not in (12543) So, (14)(25),(12)(45) is not in the normalizer of (12543). (24)(35)(12543) (24)(35) = (14325) which is not in (12543) So, (23)(45),(24)(35) is not in the normalizer of (12543). So, none of the subgroups of order 4 are in the normalizer of (12543). From above, we saw that (12)(45),(14)(25) is not in the normalizer of (13542) = (12453). We also just saw that (23)(45),(24)(35) is not in the normalizer of (14325) = (12453). We also just saw that (12)(35),(13)(25) is not in the normalizer of (13542) = (12453). So, just two more to check for (12453): (13)(24)(12453) (13)(24) = (13425) which is not in (12453) So, (12)(34),(13)(24) is not in the normalizer of (12453). (13)(45)(12453) (13)(45) = (13254) which is not in (12453)

5 So, (13)(45),(14)(35) is not in the normalizer of (12453). So, none of the subgroups of order 4 are in the normalizer of (12453). Above, we saw that (13)(45),(14)(35) is not in the normalizer of (13425) = (12354). And we saw that (23)(45),(24)(35) is not in the normalizer of (13425) = (12354). And we saw that (14)(25),(12)(45) is not in the normalizer of (14532). And we saw that (12)(34),(13)(24) is not in the normalizer of (13425) = (12354). This means we only have to check one more order- 4 subgroup for (12354). (12)(35)(12345) (12)(35) = (15432) which is not in (12354) So, (12)(35),(13)(25) is not in the normalizer of (12354). So, none of the subgroups of order 4 are in the normalizer of (12354). If the pattern holds, I should have already eliminated everything for my final subgroup of order- 5, (12435). (12)(45),(14)(25) is not in the normalizer of (14523) = (12435). (12)(34),(13)(24) is not in the normalizer of (15342) = (12435). (13)(45),(14)(35) is not in the normalizer of (13254) = (12435). (23)(45),(24)(35) is not in the normalizer of (13254) = (12435). Looks like I got out of too many checks with (12)(35),(13)(25). So here goes: (12)(35)(12435)(12)(35) = (14532) which is not in (12435). So, (12)(35),(13)(25) is not in the normalizer of (12435). So, none of the subgroups of order 4 are in the normalizer of (12435). That s all six of them! So, not only are there no normal subgroups of order 5, we also know that none of the subgroups of order 5 has a subgroup of order 4 in its normalizer. This means no semi- direct product of order 20. So there are no subgroups of order 20. This means that we know for sure we have all the subgroups! Now, it is all down hill! Our order- 6 subgroups all contain only one subgroup of order 3 (generated by a 3- cycle). We know from earlier that we can form a conjugate of any of these 3- cycles and leave the subgroup generated by that 3- cycle. This then means that we can form a conjugate of any

6 subgroup of order 6 and leave that subgroup of order 6. So, none of the order 6 subgroups are normal. Similarly, from our work above, we know that we can conjugate an 5- cycle and get another order 5- cycle outside of the subgroup generated by the original 5- cycle. Since our subgroups of order 10 are all isomorphic to D10 which contains only one subgroup of order 5, we then know that we can form a conjugate of any subgroup of order 10 that is a different subgroup of order 10. That leaves us to consider our 5 subgroups that are isomorphic copies of A4. So, let s eliminate those! These are easy, all I have to do is get the missing number to show up. Note that (125)(123)(152) = (253) which is not in (12)(34),(123). And (124)(123)(142) = (243) which is not in (12)(35),(123). And (123)(124)(132) = (234) which is not in (12)(45),(124). And (123)(134)(132) = (142) which is not in (13)(45),(134). And (123)(234)(132) = (143) which is not in (23)(45),(234). Bam! All subgroups have failed the test of normality. And therefore, A5 is simple!