# Math Help - n-divisible, n-torsionfree

1. ## 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.

2. 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.

3. 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.

4. How do I do part b) with the snake lemma [or another way]?

5. 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.

6. 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).

7. Originally Posted by zelda2139
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!

8. Originally Posted by zelda2139
(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 $c=g(b)=gf(a')=0. \ \Box$