I'm trying to prove the fundamental theorem of finitely generated abelian groups. But I cannot an argument involving free abelian group; that is, avoid using the following proposition:
(Don't-Use-This-) Proposition. Let $F$ be a free abelian group and $G$ its subgroup. Then there is a basis $x_1 ,\, \cdots,\, x_n$ of $F$, a positive integer $r$, and positive integers $d_1,\, \cdots,\, d_r$ such that $d_1 | d_2 | \cdots | d_r$ and $\{d_i x_i \}$ is a basis of $G$.
Avoid the above. This is the request of the exercise 2 in section II-2, Algebra by Hungerford, I guess.
The exercise asks first to prove and use the following observation, which I've already done.
Obsevation. Let $G$ be a finite abelian group (not 'finitely generated') and $x$ be an element of maximal order. Then the cyclic subgroup $\langle x \rangle$ of $G$ is a direct summand of $G$.
So the last thing I have to do is completing the proof of the FTFGAG with the above observation. But to do that, I have to decompose a finitely generated abelian group $G$ into two pieces: torsion part and torsion-free part, because the observation works only for finite groups. So my question is:
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Without any aid of the above Don't-Use-This-Proposition, how to decompose a fin. gen. abelian group into the torsion and torsion-free part?
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Or, is it the must? Must I rely on the D-U-T-P?
Best Answer
Presumably you know how to prove that a torsion-free FGAG is isomorphic to ${\mathbb Z}^n$ for some $n$?
The torsion subgroup $T(G)$ of a FGAG is finite, so you can apply the result you have proved to show that $T(G)$ is a direct sum of cyclic groups.
Then $G/T(G)$ is torsion-free, and you can find a complement $C \cong {\mathbb Z}^n$ of $T(G)$ in $G$ and then $G = T(G) \oplus C$, and you are done.