Jump to content

SimonLee

Members
  • Posts

    5
  • Joined

  • Last visited

Retained

  • Lepton

SimonLee's Achievements

Lepton

Lepton (1/13)

10

Reputation

  1. Suppose R is a PID which is not field, M is a finitely generated module on R. prove that if for any prime p, M/pM is cyclic R/pR module, then M is cyclic. I am just trying using the uniqueness of the structure theorem about finitely generated module over pid...but don't know how to connect M with M/pM. Any help would be appreciated. Simon
  2. Thx a lot! I am a little confused about the reason {N,G/N} = {M,G/M} implying that G/M=N, and G/N=M , why we cannot get G/N isomorphic to G/M, M isomorphic to N? And I have a stupid question, the intersection of N and G/N would be always trivial or not? If we can get G= M(semi-direct product)N and G =N(semidirect product)M,then since both M and N are normal subgroups, so the semi-direct products are direct,that's due to the definition. And I am also thinking of using the facts: M N are bothe maximal normal subgroups, (We can gurantee here that M and N are both maximal, but whether the intersection is trivial? ) then G = MN
  3. Here is the question: If a group G has a composition series of length two, then for any two distinct normal subgroups M and N, G= M x N..How to see that? what I am thinking is whether we can get G>M>1 is a composition series.(every composition series should have length two from Jordan Holder) Thanks a lot! Simon
  4. Any help on the following problem would be appreciated: Consider domain R= Q + x^2 Q[x] (polynomial without linear term) , prove that for every f, R/fR is a finite dimensional vector space on Q. and prove that every prime ideal in R is maximal~ Thanks Simon
  5. group of order 540 is simple~~ I get from sylow's theorem that r5 = 36, r3= 10, r2 = 45 ..I cannot get contracdiction with counting elements... Any help would be appreciated~ Simon
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.