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Let $A$ be a finitely generated abelian group of rank $r$. The rank of the abelian group $A$ is the number of copies of $\mathbb Z$. Let $T$ be the torsion subgroup of $A$. Show that $\frac{A}{T(A)}\cong\mathbb Z^r$.

I don't know if it helps but I've managed to show that all the non-zero elements of $T(A)$ have infinite order.

I'm guessing some usage of the FTFAG will bring out the isomorphism but I don't know how to get rid of the $\frac{\mathbb Z}{n \mathbb Z}$ bits.

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I'm confused. What is your definition of "rank $r$"? If you write $A\simeq T(A)\times \mathbb{Z}^r$, isn't this number by definition the rank in which case there isn't anything to show? – Matt Apr 28 '13 at 20:40
How have you managed to shown that the elements of $T(A)$ have infinite order?! The torsion subgroup is precisely those elements of finite order in the group... – Warren Moore Apr 28 '13 at 20:49
$\mathbb{Q}$ is a rank one torsion free abelian group, so the statement is generally false. It is true for finitely generated abelian groups. – egreg Apr 28 '13 at 20:50
It is true that $T(A)$ is finite, but I don't see why this is relevant to the problem. If it is finitely generated, then by the structure theorem $A\simeq T(A)\times \mathbb{Z}^r$, so killing $T(A)$ just leaves you with $\mathbb{Z}^r$. – Matt Apr 28 '13 at 20:58
@DonAntonio I know it's not free! It's torsion-free. Probably I should have used a hyphen. – egreg Apr 28 '13 at 22:53
up vote 1 down vote accepted

To say that the rank of an abelian group is "the number of copies of $\mathbf{Z}$" doesn't quite make sense for abelian groups which are not finitely generated. The rank of an arbitrary abelian group $A$ is by definition the $\mathbf{Q}$-dimension of the $\mathbf{Q}$-vector space $A\otimes_\mathbf{Z}\mathbf{Q}$. As egreg points out in the comments, $\mathbf{Q}$ is an abelian group of rank $1$ without torsion and it is not free of rank one because it is not finitely generated. So it is not true that an abelian group of rank $r$ is free of rank $r$ after taking the quotient modulo the torsion subgroup. For finitely generated abelian groups this is true, and follows, for example, from the structure theorem for finitely generated abelian groups (once one verifies that the number of copies of $\mathbf{Z}$ appearing in a decomposition of a finitely generated abelian group into a direct sum of cyclic groups is equal to the rank as defined above).

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Thanks. I'll edit to include that $A$ should be finitely generated. – Haikal Yeo Apr 28 '13 at 21:09

You can use the following:

according to Fuchs, Theorem 15.5 (which holds in some way even for modules over PID, I suppose) a group is finitely generated iff it is a finite direct sum of cyclic groups. So the torsion part, $T(A)$, is finitely generated as well, i.e. bounded. $T(A)$ is always pure and pure bounded subgroups are direct summands, so it has a direct complement, which is torsion-free (every torsion element is in $T(A)$) and finitely generated, hence again direct sum of cyclic groups, this time each of them is isomorphic to $\mathbb{Z}$, so $C\simeq\mathbb{Z}^n$. You get: $$A/T(A) \simeq T(A) \oplus C/T(A) \simeq C/(T(A) \cap C) \simeq C/0 \simeq C$$ and you're done.

P.S. The elements in $T(A)$ MUST be of FINITE order!

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