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Let $A$ be a finitely generated Abelian group. Let $tA$ denote the torsion subgroup. Prove that $A$ has a subgroup isomorphic to $A/tA$.

I know that $A/tA$ is torsion free, so my thinking so far has been to take the non-torsion elements of $A$ and consider the subgroup generated by them. Then show that subgroup and $A/tA$ are isomorphic. (I'm doing all this in additive notation, just so you know.)

Let $A'=\{a\in A\,|\, na\neq 0 \text{ for all } n\in \mathbb{Z}^+\}$ and consider the subgroup $B=\langle A'\rangle$. Since $A$ is finitely generated, so is $B$. Let the generators be $\{a_1, \ldots , a_n\}$.

Restrict the natural projection $\pi: A\to A/tA$ to a map $\varphi: B\to A/tA$. Since $\pi$ is a well-defined homomorphism, so is $\phi$.

Now I'd like to show that $\varphi$ is bijective, but I'm a little stuck. First of all, I haven't convinced myself it's even true. Secondly, for injectivity, does it suffice to only consider two generators $a_i$ and $a_j$ and suppose their images are distinct and go from there? Or do I have to take two arbitrary elements of $B$?

Any help would be great, thanks!

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you could show that a finitely-generated torsion-free abelian group is free abelian. Then you can just retract the projection. –  Arturo Magidin Nov 13 '10 at 21:54
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the sum of two non-torsion elements need not be non-torsion (or the identity). –  Qiaochu Yuan Nov 13 '10 at 21:58
    
@Arturo: Can you elaborate on what you mean by retracting the projection? –  Bey Nov 14 '10 at 1:38
    
Maybe I meant a section... If $f\colon A\to B$, then $g\colon B\to A$ such that $f\circ g = \mathrm{id}_B$ is a section, and $f$ is a retract of $g$. Any surjective map onto a free group can be "retracted" (made into a retract; you are "taking it back" in a sense). If $\pi\colon A\to A/tA$ is the projection, and you find $f\colon A/tA \to A$ with $\pi\circ f = \mathrm{id}_{A/tA}$, then $f$ must be one-to-one, giving you the desired subgroup of $A$. And every surjective map onto a free group is always a retract. –  Arturo Magidin Nov 14 '10 at 3:49
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3 Answers

up vote 4 down vote accepted

In fact more is true. Every finitely generated abelian group is a direct sum of its torsion part and the quotient by its torsion part. The structure theorem of finitely generated abelian groups is even stronger than this. You can find a proof of the weaker result in Peter May's notes (Theorem 1.11) http://www.math.uchicago.edu/~may/TQFT/Lecture2.pdf

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I said the classification is a stronger result and I have provided a link for the weaker result (that every finitely generated abelian group is a direct sum of it's torsion subgroup and the quotient by the torsion subgroup). It does NOT use the classficiation theorem and has been proved from first principles. –  Timothy Wagner Nov 13 '10 at 22:32
    
My apologies; I misread your answer. –  Qiaochu Yuan Nov 13 '10 at 23:58
    
@ Timothy: "Every finitely generated abelian group is a direct sum of its torsion part and the quotient by its torsion part." Using Arturo's hint above I've proved this (with some theorems we've covered in class). But how does that imply that A has a subgroup isomorphic to A/tA ? –  Bey Nov 14 '10 at 3:32
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@Bey: Since $A\cong tA \oplus A/tA$ the image of $A/tA$ in $A$ under this isomorphism is the required subgroup. –  Timothy Wagner Nov 14 '10 at 4:21
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Here is the way I hinted at the solution: proving first that a finitely generated torsion-free abelian group is free, then use the fact that free groups are projective.

The argument for the first part is that used by Lang in his Algebra: suppose $A$ is a finitely generated torsion-free abelian group; we may assume it is nonzero. Pick $x_1,\ldots,x_n\in A$ that are maximal with respect to the property that if $\alpha_1x_1+\cdots +\alpha_nx_n=0$, then $\alpha_i=0$ for all $i$. Let $B$ be the subgroup generated by the $x_i$; note that $B$ is free abelian.

Let $y_1,\ldots,y_k$ be a generating set for $A$. For each $i$, there exists an integer $d_i\neq 0$ and integers $\beta_1,\ldots,\beta_n$, not all zero, such that $d_iy +\beta_1x_1 + \cdots +\beta_n x_n=0$; that is, $d_iy_i\in B$. Letting $d=\mathrm{lcm}(d_1,\ldots,d_k)$, we have that $dA$ is contained in $B$. Since $A$ is torsion free, the map $A\to B$ given by $x\mapsto dx$ is an injection, so $A$ is isomorphic to a subgroup of a free abelian group, hence $A$ is free abelian.

Thus, if $A$ is a finitely generated group, and $tA$ is its torsion group, we know that $A/tA$ is finitely generated and torsion-free, hence free. Let $x_1,\ldots,x_k$ be a basis for $A/tA$. For each $i$, let $a_i\in A$ be any element that maps to $x_i$ under the canonical projection $\pi\colon A\to A/tA$. Since $A/tA$ is free, we have a (unique) group homomorphism $\rho\colon A/tA\to A$ that maps $x_i$ to $a_i$. Since $\pi\circ\rho = \mathrm{id}_{A/tA}$, then $\rho$ must be one-to-one, hence $A/tA$ is isomorphic to its image under $\rho$ inside of $A$. Thus, $A$ has a subgroup that is isomorphic to $A/tA$.

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Unfortunately the whole group will be generated by the nontorsion elements if the group is infinite. One thing you can do is use the classification theorem for finitely generated abelian groups which says that a finitely generated abelian group is of the form $Z_{m_1} \times ... \times Z_{m_n} \times Z \times ... \times Z$, with finitely many $Z$'s appearing on the right. In this case $tA$ is $Z_{m_1} \times ... \times Z_{m_n}$ and $A$ / $tA$ will be the $Z \times ... \times Z$ which is clearly a subgroup of $A$.

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It seems pretty clear to me that you should not be using the classification to do this. Isn't this used as a lemma in some proofs of the classification? –  Qiaochu Yuan Nov 13 '10 at 21:57
    
The point is that the statement is clear using the classification theorem. I am not trying to provide a proof from first principles. Whether or not you think I should be using the classification theorem is of no great concern to me. –  Zarrax Nov 13 '10 at 22:16
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