From: finalfntsy@aol.com (FinalFntsy) Subject: Re: Question>How to prove that no group of order 160 is simple? Date: 07 Jan 2000 11:49:19 GMT Newsgroups: sci.math Summary: [missing] The question asked was how to prove that no group of order 160 is simple. Here is how to do it. Let G be a group of order 160 = 5*2^5, and let P be a Sylow 2-subgroup of G. Then P has index 160/32 = 5 in G. Let S be the set of cosets gP of P in G and let G act on S by left multiplication. We know that gP = hP if (h^(-1))g is in P and gP intersetc hP = the null set otherwise, so this is a well defined group action. Thus there is a homomorphism from G into S_5, the permuation group on 5 letters. The image of G in S_5 is nontrivial because for g in G and g not in P, the coset P maps to gP which is not P, and at least on of the elements of g permutes the cosets in S nontrivially. Thus there is a homomorphism from G into S_5 whose kernel is a proper normal subgroup of G. Because |G| = 160 > 120 = |S_5|, this kernel must also be nontrivial. Thus G has a proper, nontrivial normal subgroup, and G is not simple. ============================================================================== From: spamless@nil.nil Subject: Re: Question>How to prove that no group of order 160 is simple? Date: 11 Jan 2000 09:27:42 -0500 Newsgroups: sci.math ÀÌ»ó¿± wrote: > Can anyone help me solving this? Let S be the collection of Sylow-p groups of G and have cardinality n. Let the elements of S be s_1, s_2, s_3, ..., s_n. For g in G, let F(g)=(gs_1g^-1,gs_2g^-1,....,gs_ng^-1), a reordering of S. F is a homomorphism from G to Symmetric(n) (let me use Symm(n) for this group of n! elements). When can G be simple? First, either G is a p-group or n>1 (else the single Sylow-p group would be a normal subgroup). If n>1, then the kernel of F must be either all of G or trivial. It cannot be all of G, for there is an element g in G where gs_1g^-1=s_2. So F must be an isomorphism and thus the order of G (|G|) must divide n! Or ... for G to be simple, if there are n Sylow-p groups, then |G| must divide n! Consider a group, G, of order 160=32*5. Can it be simple? How many Sylow-2 groups are there? Well, this is not a 2-group, so there must be 5. Then 160 would have to divide 5!=120. It doesn't. --- This also works for groups of order 80 (80 doesn't divide 120 either!). But this proof doesn't work for groups of order 40=8*5 (40 DOES divide 120). But there can't be a simple group of order 40 since a group of such order must have 1 Sylow-5 group (a normal subgroup) (Any group which is simple and has order 5*2^j must have j>=4 so it can have more than 1 Sylow-5 group). ============================================================================== From: spamless@Nil.nil Subject: Re: Question>How to prove that no group of order 160 is simple? Date: 11 Jan 2000 10:50:16 -0500 Newsgroups: sci.math spamless@nil.nil wrote: (Any group which is > simple and has order 5*2^j must have j>=4 so it can have more than 1 > Sylow-5 group). And in that case, they can't be simple, since then 5*2^j does not divide 5!=120 (so there is just one Sylow-2 group)