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You are not logged in. #1 20080925 07:37:29
Simple groupsJust got this as a homework assignment: "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #2 20080925 07:46:46#3 20080925 07:50:11
Re: Simple groupsCertainly a powerful tool, but not the only thing you use. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #4 20081009 07:48:06
Re: Simple groupsFor a start, all groups of order where is an odd prime and are not simple. By Sylow, there would be a subgroup of order and such a subgroup would be normal because it has index 2. That is what I proved here: http://www.mathisfunforum.com/viewtopic.php?id=10264.This takes care of all groups with the following orders: 6, 10, 14, 18, 22, 26, 34, 38, 46, 50, 54, 58, 62, 74, 82, 86, 94, 98, 106, 118. #5 20081009 09:12:11
Re: Simple groupsIndeed, you can use Sylow's Theorem directly to prove an even more general result: "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #6 20090413 04:31:08
Re: Simple groupsYes, thats Corollary 9.25 in Humphreyss book (see this post). Now that I finally understand this, I can shoot down some more nonsimple finite groups: #7 20090413 05:47:25
Re: Simple groupsI know. Ill list all the group orders and check them one by one. When I make another breakthrough, Ill update the list. Last edited by JaneFairfax (20090516 22:02:48) #8 20090413 06:02:18
Re: Simple groupsAccording to Humphreys (pp 4647) there are five nonisomorphic groups of order 8. Three of these are Abelian, being products of cyclic groups. Since they are Abelian, all their subgroups are normal (and they will definitely have subgroups of order 2). The other two are the dihedral group and the quaternion group. In the former, the group of rotations are a subgroup of index 2. Im not familiar with properties of the quaternion group, but I do gather that there is an element of order 4. Hence the quaternion group also has a subgroup of index 2. So these two groups are not simple. So I have convinced myself that all subgroups of order 8 are not simple. Ill go and update my post above. Last edited by JaneFairfax (20090413 06:07:28) #9 20090413 13:00:20
Re: Simple groupsJane, all pgroups have a nontrivial center. That means not only are all groups of order 8 not simple, but any power of 2 (or any prime for that matter) as well. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #10 20090413 13:22:43
Re: Simple groupsAny Abelian group of order 9 will definitely have a subgroup of order 3; hence its not simple. Suppose there is a nonabelian group of order 9. Every nonidentity element must have order 3, so we can list 7 of its elements straightaway: Now, what can the element be? It cant be any of the first six elements. Suppose (so ). Then Contradiction! Hence the elements of this group must be So now the element must be one of these. It obviously cant be any of the first six or . If then cancelling gives (contradiction); (again!). Hence Wow, I have just made an amazing discovery on my own at four in the morning!! #11 20090413 13:25:24
Re: Simple groupsSorry, Ricky. I was posting when you were. #12 20090413 19:09:59
Re: Simple groups
Got it! Its Proposition 10.20 on page 94 of Humphreys! Last edited by JaneFairfax (20090413 19:19:34) #13 20090413 20:00:33
Re: Simple groupsThe Sylow 2subgroup has index 3 in , so using a result in Humphreys, there is a subgroup normal in and contained in such that 3 divides and divides 3! = 6. Hence is either 3 or 6. This ensures that is never 1 or therefore is not simple! Thank goodness for 3! being a conveniently small number. Last edited by JaneFairfax (20090414 05:00:28) #14 20090414 04:40:58
Re: Simple groupsThe Sylow non2subgroup has index 4 in , so there is a subgroup normal in and contained in such that 4 divides and divides 4! = 24. (in the case of the 108, ) is not simple. Yo, man! I hope Ricky is appreciating my hard work! Last edited by JaneFairfax (20090512 18:58:48) #15 20090414 10:11:37
Re: Simple groupsFrom post #13:
Can you post this result? I know of a similar result, but one must assume G is simple to use it (and hence you'd be going by contradiction).
Certainly. But perhaps it is wiser to read more on Sylow first before trying to prove more of these. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #16 20090415 00:59:15
Re: Simple groups
The result is Corollary 9.23 in the book by Humphreys:
Well, yes. Ive just started reading the chapter on the Sylow theorems in Humphreys. So now I know a few more tricks: The number of Sylow subgroups of is congruent to 1 modulo and divides . If the Sylow subgroup is unique, then it is normal in . Now Ill go away and work on these; in a day or two, Ill come back and give the results. #17 20090415 01:30:38
Re: Simple groupsOkay, lets start at the beginning. Let be a group, a subgroup and the set of all left cosets of in . Any element induces a permutation by for each left coset . Yes, its well defined, and its bijective. Moreover, forms a group isomorphic to a subgroup of the symmetric group . Now define a homomorphism by . Yup, its a homomorphism all right. The kernel of is this peculiar subgroup of : This is the N that is referred to in Humphreyss Corollary 9.23. By imposing special properties on the subgroup , you can have a normal subgroup with special properties. Corollary 9.23 is obtained by imposing the condition on . As an aside, you could also impose the condition that be the trivial group . The result is Cayleys theorem: Every group is isomorphic to a subgroup of a symmetric group. This is Corollary 9.24 in Humphreys. Last edited by JaneFairfax (20090415 23:50:16) #18 20090415 02:11:26
Re: Simple groupsAll right, lets go one step further and prove Corollary 9.23. Since is normal in and is a subgroup of , is a (normal) subgroup of and . divides . The second assertion is proved by the fact that is isomorphic to the image of in the previous post, which is a subgroup of , which is isomorphic to a subgroup of . divides . Well, Ricky, I hope you are checking everything I say and ascertaining that it makes sense. You have to, because Im going through a bottle of rosι wine while posting all this stuff so I could easily have messed up something here and there. Last edited by JaneFairfax (20090415 02:16:57) #19 20090415 23:45:14
Re: Simple groupsThese are groups of orders where is a prime and . I will show that the Sylow subgroup of any such group is unique. The number of Sylow subgroups is of the form . Each of these is cyclic of prime order; therefore distinct Sylow subgroups have only the identity in common. So the total number of elements in the union of all the Sylow subgroups is . This cannot exceed the total number of elements in . Hence ; in other words, there is only 1 Sylow subgroup! This must be normal in since it is equal to all its conjugates in ; therefore is not simple. Well! This is the first time I am diligently making use of the Sylow tools; previously Ive only used Humphreyss Corollay 9.23 with only the knowledge that Sylow subgroups exist. Im fairly happy with myself for this. Last edited by JaneFairfax (20090416 05:20:52) #20 20090416 19:28:57
Re: Simple groupsFor 90, I made a mistake so forget it for the time being. In the other cases, the Sylow subgroup for the max prime divisor is unique, therefore normal, as can be ascertained by patiently checking each individual case. Last edited by JaneFairfax (20090513 04:07:51) #21 20090423 05:24:59
Re: Simple groupsIncidentally its been pointed out to me that my result in #19 could also be proved by considering the fact that divides . If then and so would not divide . Hence divides only if . #22 20090423 17:50:58
Re: Simple groupsIf then has 1 or 6 Sylow 5subgroups and 1 or 10 Sylow 3subgroups, each Sylow subgroup being cyclic of prime order. If there are 6 Sylow 5subgroups, the union of these subgroups has elements; if there are 10 Sylow 3subgroups, their union will have elements, all distinct from the elements of the other union except the identity. Since is only 30, therefore cannot have 6 Sylow 5subgroups and 10 Sylow 3subgroups at the same time; hence must have either a unique (therefore normal) Sylow 5subgroup or a unique (∴ normal) Sylow 3subgroup. If then has 1 or 15 Sylow 7subgroups and 1 or 21 Sylow 5subgroups (each Sylow subgroup being cyclic of prime order). By a similar argument to the have, we find that must have either a normal Sylow 7subgroup or a normal Sylow 5subgroup. If then has 1 or 8 Sylow 7subgroups (cylic of order 7). If 8, then the union of the the Sylow 7subgroups has 48 nonidentity elements. This leaves 56 − 48 = 8 other elements in the group (including the identity), which must then make up the unique Sylow 2subgroup. Hence must have either a normal Sylow 7subgroup or a normal Sylow 2subgroup. Last edited by JaneFairfax (20090513 04:03:28) #23 20090424 14:51:19
Re: Simple groupsPost #20 is incorrect for groups of order 90. It is possible to have six Sylow 5subgroups. I'll probably be checking the rest over once the semester is... "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #24 20090424 23:03:47
Re: Simple groupsOops, youre right. I was a bit careless there. Last edited by JaneFairfax (20090512 22:25:18) #25 20090512 19:22:42
Re: Simple groupsThe Sylow 2subgroup has index 5 and so by Humphreyss Corollary 9.23 there is a normal subgroup with divisible by 5 and dividing 5! = 120. must also divide 80. is not simple. 