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Math Help - n-divisible, n-torsionfree

  1. #1
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    n-divisible, n-torsionfree

    Let n be a nonzero integer. An abelian group A is called n-divisible if for every x \in A, there exists y \in A such that x=ny. An abelian group A is called n-torsionfree if nx=0 for some x \in A implies  x=0. An abelian group A is called uniquely n-divisible if for any x \in A, there exists exactly one y \in A such that x=ny.
    Let \mu_n : A \rightarrow A be the map \mu_n(a)=na
    (a) Prove that A is n-torsionfree iff \mu_n is injective and that A is n-divisible iff \mu_n is surjective.
    (b) Now suppose 0 \rightarrow A  \overset{f}{\rightarrow}  B   \overset{g}{\rightarrow} C \rightarrow 0    is an exact sequence of abelian groups. It is easy to check that the following diagram commutes:
    Suppose that B is uniquely n-divisible. Prove that C is n-torsionfree if and only if A is n-divisible.
    Attached Thumbnails Attached Thumbnails n-divisible, n-torsionfree-2.png  
    Last edited by zelda2139; December 11th 2008 at 10:11 PM.
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  2. #2
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    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. Any ideas on part a) or b) would be great, I never used the snake lemma to prove something yet.
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  3. #3
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    You're right for part a), you just need to apply the definitions. For instance,

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

    Same thing for the second iff.
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    How do I do part b) with the snake lemma [or another way]?
    Last edited by zelda2139; December 13th 2008 at 10:36 AM.
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  5. #5
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    For part b):
    You need to take the kernels and cokernels of the vertical mappings and apply the snake lemma. Also, write down the exact sequence to use the snake lemma. I am not sure how to actually do that though.
    Last edited by GaloisTheory1; December 13th 2008 at 02:43 PM.
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  6. #6
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    Here is how to do the second part of part a):

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


    I am still not sure how to do part b).
    Last edited by GaloisTheory1; December 13th 2008 at 09:37 PM.
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  7. #7
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    Quote Originally Posted by zelda2139 View Post
    Let n be a nonzero integer. An abelian group A is called n-divisible if for every x \in A, there exists y \in A such that x=ny. An abelian group A is called n-torsionfree if nx=0 for some x \in A implies  x=0. An abelian group A is called uniquely n-divisible if for any x \in A, there exists exactly one y \in A such that x=ny.
    Let \mu_n : A \rightarrow A be the map \mu_n(a)=na
    (b) Now suppose 0 \rightarrow A  \overset{f}{\rightarrow}  B   \overset{g}{\rightarrow} C \rightarrow 0    is an exact sequence of abelian groups. It is easy to check that the following diagram commutes:
    Suppose that B is uniquely n-divisible. Prove that C is n-torsionfree if and only if A is n-divisible.
    I still do not know how to solve (b) using the snake lemma or any other ways. help!
    Last edited by zelda2139; December 14th 2008 at 07:17 PM.
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  8. #8
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    Quote Originally Posted by zelda2139 View Post
    (b) Now suppose 0 \rightarrow A \overset{f}{\rightarrow} B \overset{g}{\rightarrow} C \rightarrow 0 is an exact sequence of abelian groups. It is easy to check that the following diagram commutes:
    Suppose that B is uniquely n-divisible. Prove that C is n-torsionfree if and only if A is n-divisible.
    you don't need snake lemma. it's pretty straightforward: suppose first that C is n-torsionfree and a \in A. then f(a) \in B, and therefore there exists a (unique) b \in B such that f(a)=nb.

    thus: 0=gf(a)=g(nb)=ng(b). hence g(b)=0, because C is n-torsionfree. so b \in \ker g=\text{Im}f. therefore b=f(a'), for some a' \in A. thus f(a)=nb=f(na'), which gives us a=na',

    since f is injective. so A is n-divisible and we're done. conversely: suppose that A is n-divisible and nc=0, for some c \in C. we have to prove that c=0. we have g(b)=c, for some b \in B,

    because g is surjective. so g(nb)=0, and hence nb \in \ker g = \text{Im}f, i.e. there exists a \in A such that f(a)=nb. now since A is n-divisible, there exists a' \in A such that a=na'. therefore

    n(f(a') - b)=0=n \cdot 0, which gives us f(a')=b, because B is uniquely n-divisible. hence c=g(b)=gf(a')=0. \ \Box
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