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You are not logged in. #26 20090513 01:20:27
Re: Simple groupsPost #19 can be simplified quite a bit. You know that the number of Sylow psubgroups, call this n_p, must be congruent to 1 modulo p. Further, n_p must divide k, where our group is of order p*k with p > k. But if n_p is not 1, then it must be greater than p, hence greater than k, and therefore n_p can not divide k. "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..." #27 20090513 01:24:17
Re: Simple groupsFrom post #20
You can not conclude this from direct application of the Sylow theorems. There could be 10 Sylow 3subgroups. Even if you could, simply stating this is not sufficient. "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..." #28 20090513 01:36:25
Re: Simple groupsPost #22 has the typo But is fine otherwise. "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..." #29 20090513 04:12:59
Re: Simple groupsOops, I did it again for 90. Oh well, back to square one. #30 20090513 07:31:21
Re: Simple groupsIf you want some help, there is a theorem which is quite useful for both Sylow theory and in general. I've scanned (briefly) through what I could get my hands on in Humphrey, but could not find this theorem. "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..." #31 20090516 18:57:21
Re: Simple groupsOkay, this trick is adapted from something I’ve just noticed on p.161 of Humphreys, which is devoted to a much different problem (namely the proving that every group of order 36 is soluble). There are 1 or 10 Sylow 3subgroups, each of order 9. If 10, the union of all the Sylow 3subgroups contains at most elements. If there are really that many, then there cannot be 6 Sylow 5subgroups, so the Sylow 5subgroup in this case is normal. Otherwise there exist distinct Sylow 3subgroups and with a nontrivial intersection. It follows that . The set therefore contains elements. So , the subgroup generated by and , must contain at least 27 elements. Its order also must be divisible by 9 and must divide 90. Hence . If 45, then is a normal subgroup, being of index 2. Otherwise, it is the whole group ; then I claim that is a normal subgroup of . For and are both Abelian (see post #10 above) and so is normal in both of them; hence must be normal in the subgroup generated by and . Whew! That took a while to get through. #32 20090516 20:11:37
Re: Simple groupsSuch a group has 1 or 4 Sylow 3subgroups, each of order 9. Suppose 4 and let be one of them. Then and so by Humphreys’s corollary, there is a normal subgroup of contained in such that 4 divides and divides 4! = 24. showing that is not simple. Wow, it really is simple when you know how. Last edited by JaneFairfax (20090516 23:31:42) #33 20090516 20:40:59
Re: Simple groupsAnd finally: There are 1 or 7 Sylow 2subgroups. If 7, the union of all the Sylow 2subgroups contains at most elements. If this be the case, the remaining 6 elements together with the identity must make up the unique Sylow 7subgroup. Otherwise let and be distinct Sylow 2subgroups. Then and so has more than 16 elements. As must have order divisible by 16 and dividing 112, this means . If then has index 2 in and , so is normal in and and hence . This leaves . Suppose all the 7 Sylow 2subgroups have pairwise intersections of 4 elements. Then the minimum number of elements in their union is when the same 4 elements are in all their intersections, which would be . This would mean there could not be 8 Sylow 7subgroups and so the Sylow 7subgroup must be normal. Whee! I did it! Well, at least I hope I have. Last edited by JaneFairfax (20090516 21:51:35) #34 20090517 00:28:54
Re: Simple groupsAnd now, how about a bonus extra? I’ve just seen two proofs of this on the Web. Here is my own adaptation of them. Note that it assumes the knowledge that is simple. Suppose there is a simple group of order 120. The number of Sylow 5subgroups must be 6. If is one of them, then and so there is a homomorphism with . (Indeed is none other than the in my oftquoted Humphreys corollary. ) As is simple, . Hence is injective and so there is a simple subgroup of of order 120. As , we have and so by simplicity of either or . The former is impossible as it would imply that the number of elements of the form for would be – way above the order of . Thus . Then and so there is a homomorphism with . By simplicity of , meaning that is injective and leading to the utter balderdash that is isomorphic to a subgroup of . This proves by contradiction that cannot be simple. In fact, this method can also be used to show that groups of order 90 are not simple as well. However, the method I used has the advantage of not assuming that the alternating group of degree 6 is simple. The fact that is simple for is established in several stages. First it is shown that that is simple. After a couple of other results are proved, the job is completed using mathematical induction. The proof of the simplicity of is readably done in not more that two pages by Humphreys. Humphreys first demonstrates that has precisely five conjugacy classes, one of which is the singleton class containing just the identity. Any normal subgroup of would have to be a union of this trivial conjugacy class with none, some or all of the other conjugacy classes. By determining the sizes of all the conjugacy classes, Humphreys is able to show that the number of elements in any union of the trivial class with some but not all of the other classes cannot be a divisor of 60, thus demonstrating by Lagrange that cannot have any nontrivial proper normal subgroup. Last edited by JaneFairfax (20090517 00:36:04) #35 20090517 02:43:28
Re: Simple groupsNow that you're done, the theorem I was speaking of was: Using this, here is how I did it. First I proved this lemma: Let p be a prime, and G = p^k * m, where it may be that p  m. If G does not divide m!/2, then G is not simple. Now using 120 as an upper bound for G, this may be now used to conclude statements: If 2^4 divides G and G < 120, then G is not simple. If 3^3 divides G and G < 120, then G is not simple. If 5^2 divides G and G < 120, then G is not simple. If 11 divides G and G < 120, then G is not simple. If 13 divides G and G < 120, then G is not simple. If 17 divides G and G < 120, then G is not simple. If 19 divides G and G < 120, then G is not simple. These divisibility conditions will get 22 group orders. We can squeeze more out of it if we forgo the use of the upper bound and just use a particular order. This will get groups of order 12, 18, 20, 24, 28, 36, 40, 42, 45, 74, 92, 98, 116. Each argument is just finding the right p^k so that G = p^k * m where p^k does not divide m!/2. Most of the rest are prime powered or groups of order pq (with p ≠ q), all of which of course not simple. This only left 8 groups which were rather manageable. On each of these, either the counting arguments or a direct application of the theorem I gave above worked. That is, on all except 90. Here was my proof for 90: Let G be a group of order 90, and assume G is simple. The number of Sylow 5subgroups must divide 90/5 = 18, and is congruent to 1 modulo 5. As G is simple, we conclude this number must be 6. Let H be a Sylow 5subgroup, and so [G : N(H)] = 6. Thus, we may identify G as a subgroup of A_6. But now A_6 = 360, G = 90, and so [A_6, G] = 4, with A_6 being a simple subgroup. Thus, A_6 may be identified with a subgroup of A_4, a contradiction. "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..." #36 20090517 04:35:54
Re: Simple groupsYeah, I wasn’t sure if I should really use the simplicity of the alternating groups. I didn’t know what counted as “heavy machinery”. You are obviously accustomed to a lot of heavyduty stuff in group theory whereas I am only picking up from where I left off years ago, so what may be tough going for me may be feather light for you. #37 20090517 06:12:32
Re: Simple groups
Heavy machinery to me would be the theorem of Burnside and Feit–Thompson. With those, the task becomes rather trivial.
I'd be impressed if we did all groups of order up to 1000. "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..." #38 20090609 07:15:21
Re: Simple groupsI have a question for JaneFairfax or anyone who want it to answer, in the #22 in the case that the order of G is 56 why the union of the the Sylow 7subgroups has 48 nonidentity elements? i mean why the intersection is only the indentity? sorry maybe that is very easy to see but im not quite familiar with this definitions.. #39 20090609 07:26:33
Re: Simple groupsIn a group G of order 56, the Sylow 7subgroups have order 7. Now let's assume we have two Sylow subgroups, P and Q. Their intersection, call it I, has to be not only a subgroup of G, but quite clearly a subgroup of P as well. By Lagrange's theorem, the order of I must divide the order of P. But P's order is 7. So either we have that the order of I is 1 (the identity) or 7 (all of P). "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..." #40 20090620 00:27:54
Re: Simple groupsThank you! the intersection can't be all of P because in that case the intersection of P and Q must be P, this is, P C Q and that it's a contradiction since P is a Sylow Psubgroup a maximal psoubgroup.......????????????? #41 20090620 21:56:45
Re: Simple groupsYou're certainly on the right track. But when you have that the intersection contains all of P, and hence all of Q, you actually get that P and Q are the same exact two subgroups. Since we started out with two different subgroups, obviously this can't happen. "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..." 