Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

N-divisible, n-torsionfree

  1. Dec 8, 2008 #1
    Let [tex]n[/tex] be a nonzero integer. An abelian group [tex]A[/tex] is called [tex]n[/tex]-divisible if for every [tex]x \in A[/tex], there exists [tex]y \in A[/tex] such that [tex]x=ny[/tex]. An abelian group [tex]A[/tex] is called [tex]n[/tex]-torsionfree if [tex]nx=0[/tex] for some [tex]x \in A[/tex] implies [tex] x=0[/tex]. An abelian group [tex]A[/tex] is called uniquely [tex]n[/tex]-divisible if for any [tex]x \in A[/tex], there exists exactly one [tex]y \in A[/tex] such that [tex]x=ny[/tex].
    Let [tex]\mu_n : A \rightarrow A[/tex] be the map [tex]\mu_n(a)=na[/tex]
    (a) Prove that [tex]A[/tex] is [tex]n[/tex]-torsionfree iff [tex]\mu_n[/tex] is injective and that [tex]A[/tex] is [tex]n[/tex]-divisible iff [tex]\mu_n[/tex] is surjective.
    (b) Now suppose [tex]0 \rightarrow A \overset{f}{\rightarrow} B \overset{g}{\rightarrow} C \rightarrow 0 [/tex] is an exact sequence of abelian groups. It is easy to check that the following diagram commutes: [see attachment]
    Suppose that [tex]B[/tex] is uniquely [tex]n[/tex]-divisible. Prove that [tex]C[/tex] is [tex]n[/tex]-torsionfree if and only if [tex]A[/tex] is [tex]n[/tex]-divisible.
     

    Attached Files:

    • 1.PNG
      1.PNG
      File size:
      2.4 KB
      Views:
      67
  2. jcsd
  3. Dec 8, 2008 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's hard to be helpful if I have no idea where you're stuck....
     
  4. Dec 10, 2008 #3
    So I am now thinking to use the snake lemma for part b). However, I am a little confused on how to do this. Also, for part a) these statements seem a bit trivial, I will work on those proofs using the definitions. Do you have any ideas on part a) or b), I never used the snake lemma to prove something yet.
     
  5. Dec 10, 2008 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I agree, (a) looks easy. I think it's just a matter of writing out the definitions, as you say.

    For (b) you have the right idea -- in the two cases (assuming C n-torsion-free and assuming A n-divisible), what did you compute as the sequence you get from the snake lemma?
     
  6. Dec 11, 2008 #5
    So here is what I have for the first part of a).

    [tex]A[/tex] is [tex]n[/tex]-torsionfree
    [tex]\Leftrightarrow \forall x\in A\ (nx=0 \Rightarrow x=0)[/tex]
    [tex]\Leftrightarrow \forall x\in A\ (\mu_n(x)=0 \Rightarrow x=0)[/tex]
    [tex]\Leftrightarrow \mu_n[/tex] is injective

    I am confused now on how to do the second part of a).
     
  7. Dec 12, 2008 #6
    How do I do the second part with the snake lemma? I do not know how to do this part.

    Here is the picture:
    Picture
     
    Last edited: Dec 12, 2008
  8. Dec 12, 2008 #7

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Doesn't the same idea work? What went wrong?

    Can you be more precise? Is it... that you don't know how to write down the exact sequence given by the snake lemma? Is it... that you don't know what to do with the sequence one you have it? Is it... something else?
     
  9. Dec 13, 2008 #8
    So, I need to take the kernels and cokernels of the vertical mappings and apply the snake lemma. Also, I need to write down the exact sequence to use the snake lemma. I am not sure how to actually do this. I never used the snake lemma to prove something before. I guess I just need to know how to start.
     
  10. Dec 13, 2008 #9
    Here is how to do the second part of part a):

    [tex]\mu_n \text{ is surjective}[/tex]
    [tex]\Leftrightarrow \forall y\in A,\, \exists x\in A \, (\mu (x) = y)[/tex]
    [tex] \Leftrightarrow \forall y\in A,\, \exists x\in A\, (nx = y)[/tex]
    [tex] \Leftrightarrow \text{ A is n-divisible}[/tex]

    For part b):
    I am still not sure how to do part b).
     
  11. Dec 13, 2008 #10

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't see what the problem is. I presume you know what the 6 groups are -- you even have concrete descriptions of many of them. (As opposed to a purely formal description, such as "[itex]\\ker \left( B \xrightarrow{\mu_n} B \right) [/itex]") I presume you even know how those groups are to be organized into an exact sequence. So I don't understand why you are not sure how to write the exact sequence.
     
  12. Dec 14, 2008 #11
    I still do not know how to solve (b) using the snake lemma or any other ways. help! :confused:
    When I use the snake lemma, how will this help me with what I am trying to show? And for both directions? I am very confused.

    just solved it. :-)
     
    Last edited: Dec 14, 2008
  13. Dec 14, 2008 #12

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    From the information you've given me, it sounds like you haven't even bothered to write down what the snake lemma tells you in this case! And if you haven't even done that, then of course you don't know how it will help!

    It's okay to try something without knowing how it will be useful. In fact, doing so is required for doing mathematics. (In fact, it's a requirement for doing research in any field!)

    So... just do it! Write down what the snake lemma tells you, and figure out everything you can from there. If you can't manage to see your way to the end of the problem, then come back here and show us how far you have gotten, and we can help you figure out what you've overlooked.
     
  14. Dec 14, 2008 #13
    I solved it Hurkyl (as I said in my post if you read it thoroughly). And no, I didn't use the snake lemma at all. One step was messing me up, but I got through it.
     
  15. Dec 15, 2008 #14
    To me it looks like an easy application of the four lemmas, no? Since
    • A is n-torsion free ⇔ μn is a monomorphism
    • A is n-divisible ⇔ μn is an epimorphism
    • A is uniquely n-divisible ⇔ μn is an isomorphism
    Moreover, it proves the stronger statement that if B is n-divisible and C is n-torsion free then A is n-divisible, and if B is n-torsion free and A is n-divisible, then C is n-torsion free. I'm pretty sure that was the point of part (a) actually: setting it up for proving (b).
     
    Last edited: Dec 15, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: N-divisible, n-torsionfree
  1. Divisibility of n by 6 (Replies: 10)

Loading...