From: jamesrheckman@aol.comnospam (Jim Heckman) Subject: Re: group of order 168 Date: 10 Apr 2001 06:29:36 GMT Newsgroups: sci.math Summary: Constructing the simple group of order 168 >From: rkbqk@yahoo.com (none) >Date: 4/9/01 11:27 AM Pacific Daylight Time >Message-id: > >How? I produced generated groups of order 3, 7 and 8, >hence the size of the group is 168n. It can't be of index >5 or 3 in A_7, so that leaves two options, either 168 >or the whole A_7. OK, but I'm curious how you know A_7 can't have subgroups of index 3 or 5. Using only the tools you mention below, I would think that just proving *that* would be pretty heavy! >I did mean GL_3(F_2), not F_7...sorry about that. >The book only covered group theory and nothing about >linear groups at the point where this problem occurs. >But everything else, like sylow theorems, etc..is available. Well, I assume you've seen Robin Chapman's post where he pulls a specific set of 7 triplets from {1,2,...,7} out of his *ss :-) and uses them to show your group can't possibly be A_7. But as I posted in a follow-up to Robin, you probably have to know at least a little permutation group theory (or some of the properties of GL(3,2)) before you can guess that triplets are the right way to go -- e.g., that just pairs probably won't, and in fact don't, work. You could also follow through on my approach below. I could help you step through that process of showing that your group must be of order 168, but it depends on knowing ahead of time something about the conjugacy classes and subgroup structure of the group -- i.e., that it's the simple group of that order. Have you already shown that if your group is of order 168, it must be simple? If so, I could take you on from there... >>From: rkbqk@yahoo.com (group of order 168) >>Date: 4/7/01 8:01 PM Pacific Daylight Time >>Message-id: >> >>Hi, >> >> I'm trying to show that the group generated by >>(1234567) and (26)(34) is of order 168. I showed that >>it must be either 168, > >How did you do that? That may give us a clue about >an easier approach than what I outline below. > >>or the whole A_7, but I can't >>figure out how to show that some even permutation isn't >>generated by these two elements (which would prove it's not >>A_7). > >How about finding a specific permutation rep which contains >these two elements, of the simple group of order 168? I.e., >explicitly construct a subgroup of A_7 that contains the two >elements -- so they can't generate a larger subgroup. The >simple group is relatively easy to construct abstractly as >G = AB, where A is S_4 and B is C_7 -- just using Sylow's >Theorems, intersections of normalizers, counting arguments, >etc. -- and then relatively easy to find degree-7 permutation >reps. > >>I posted this long ago on alt.math.undergrad but I don't think >>anyone got it, so I'll try it here. Btw...I know there's a way >>to produce an isomorphism into GL_3(F_7) > >? Do you not mean GL_3(F_2), or PSL_2(F_7)? > >>but that's not the tools the book assumes, > >What tools does the book assume? > >>so I wonder if there's another way. >>Thanks to anyone who has a suggestion. -- Jim Heckman ============================================================================== From: rkbqk@yahoo.com (rjohnson) Subject: re: group of order 168 Date: 11 Apr 2001 16:25:20 -0400 Newsgroups: sci.math --------------- OK, but I'm curious how you know A_7 can't have subgroups of index 3 or 5. Using only the tools you mention below, I would think that just proving *that* would be pretty heavy! --------------- Actually it's trivial. If we have a subgroup of index 3 or 5, than let A_7 act on this subgroup's cosets by multiplication. The kernel of this action is contained in the subgroup and is normal in A_7, so the only possiblity is for kernel to be a "1". That means A_7 is isomorphic to a subgroup of S_3 or S_5, both impossible by order considerations. -------------------- Well, I assume you've seen Robin Chapman's post where he pulls a specific set of 7 triplets from {1,2,...,7} out of his *ss :-) and uses them to show your group can't possibly be A_7. But as I posted in a follow-up to Robin, you probably have to know at least a little permutation group theory (or some of the properties of GL(3,2)) before you can guess that triplets are the right way to go -- e.g., that just pairs probably won't, and in fact don't, work. You could also follow through on my approach below. I could help you step through that process of showing that your group must be of order 168, but it depends on knowing ahead of time something about the conjugacy classes and subgroup structure of the group -- i.e., that it's the simple group of that order. Have you already shown that if your group is of order 168, it must be simple? If so, I could take you on from there... -------------- I did see Robin's approach, and your variant of it, but unfortunately, I could pull this out of my **s due to insufficient knowledge. Btw...proving that this group is simple makes use of its size, so what you suggest would be circular reasoning.