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Thread: n-divisible, n-torsionfree

  1. #1
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    n-divisible, n-torsionfree

    Let $\displaystyle n$ be a nonzero integer. An abelian group $\displaystyle A$ is called $\displaystyle n$-divisible if for every $\displaystyle x \in A$, there exists $\displaystyle y \in A$ such that $\displaystyle x=ny$. An abelian group $\displaystyle A$ is called n-torsionfree if $\displaystyle nx=0$ for some $\displaystyle x \in A$ implies $\displaystyle x=0$. An abelian group $\displaystyle A$ is called uniquely $\displaystyle n$-divisible if for any $\displaystyle x \in A$, there exists exactly one $\displaystyle y \in A$ such that $\displaystyle x=ny$.
    Let $\displaystyle \mu_n : A \rightarrow A$ be the map $\displaystyle \mu_n(a)=na$
    (a) Prove that $\displaystyle A$ is $\displaystyle n$-torsionfree iff $\displaystyle \mu_n$ is injective and that $\displaystyle A$ is $\displaystyle n$-divisible iff $\displaystyle \mu_n$ is surjective.
    (b) Now suppose $\displaystyle 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 $\displaystyle B$ is uniquely $\displaystyle n$-divisible. Prove that $\displaystyle C$ is $\displaystyle n$-torsionfree if and only if $\displaystyle A$ is $\displaystyle n$-divisible.
    Attached Thumbnails Attached Thumbnails n-divisible, n-torsionfree-2.png  
    Last edited by zelda2139; Dec 11th 2008 at 09:11 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,

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

    Same thing for the second iff.
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  4. #4
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    How do I do part b) with the snake lemma [or another way]?
    Last edited by zelda2139; Dec 13th 2008 at 09:36 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; Dec 13th 2008 at 01:43 PM.
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  6. #6
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    Here is how to do the second part of part a):

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


    I am still not sure how to do part b).
    Last edited by GaloisTheory1; Dec 13th 2008 at 08:37 PM.
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  7. #7
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    Quote Originally Posted by zelda2139 View Post
    Let $\displaystyle n$ be a nonzero integer. An abelian group $\displaystyle A$ is called $\displaystyle n$-divisible if for every $\displaystyle x \in A$, there exists $\displaystyle y \in A$ such that $\displaystyle x=ny$. An abelian group $\displaystyle A$ is called n-torsionfree if $\displaystyle nx=0$ for some $\displaystyle x \in A$ implies $\displaystyle x=0$. An abelian group $\displaystyle A$ is called uniquely $\displaystyle n$-divisible if for any $\displaystyle x \in A$, there exists exactly one $\displaystyle y \in A$ such that $\displaystyle x=ny$.
    Let $\displaystyle \mu_n : A \rightarrow A$ be the map $\displaystyle \mu_n(a)=na$
    (b) Now suppose $\displaystyle 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 $\displaystyle B$ is uniquely $\displaystyle n$-divisible. Prove that $\displaystyle C$ is $\displaystyle n$-torsionfree if and only if $\displaystyle A$ is $\displaystyle n$-divisible.
    I still do not know how to solve (b) using the snake lemma or any other ways. help!
    Last edited by zelda2139; Dec 14th 2008 at 06:17 PM.
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  8. #8
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    Quote Originally Posted by zelda2139 View Post
    (b) Now suppose $\displaystyle 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 $\displaystyle B$ is uniquely $\displaystyle n$-divisible. Prove that $\displaystyle C$ is $\displaystyle n$-torsionfree if and only if $\displaystyle A$ is $\displaystyle n$-divisible.
    you don't need snake lemma. it's pretty straightforward: suppose first that $\displaystyle C$ is $\displaystyle n$-torsionfree and $\displaystyle a \in A.$ then $\displaystyle f(a) \in B,$ and therefore there exists a (unique) $\displaystyle b \in B$ such that $\displaystyle f(a)=nb.$

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

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

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

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