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Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups of the form $$\Bbb{Z_{(p_1)^{r_1}}}\times \Bbb{Z_{(p_2)^{r_2}}}\times\dots\times \Bbb{Z_{(p_n)^{r_n}}}\times\underbrace{\Bbb{Z}\times\Bbb{Z}\dots\times\Bbb{Z}}_{\text{r times, r : betti number}}$$ where $p_i$ are primes, not necessarily distinct, and $r_i$ are positive integers. The prime powers $(p_i)^{r_i}$ are unique.

Given the Betti number equals the number of elements in some generating set in a finitely generated abelian group $G$.
To show that $G$ is free abelian, I want to show that $G$ is torsion-free.

Let $\{x_1,\dots,x_r\}$ be a generating set of $G$ and $T$ be a torsion group of $G$.
Then $G/T$ is of rank $r$ and is generated by $\{x_1T,\dots,x_rT\}$. But here we are not sure that whether $\{x_1T,\dots,x_rT\}$ forms a basis for $G/T$. Also I still can't get any relevant information to show that $G$ is torsion-free.

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You have $G\cong T\times\Bbb Z^r$ where $T$ is torsion. Then $G$ has $r$ generators if there's an onto homomorphism $\phi:\Bbb Z^r\to G$. We might as well assume that $G=T\times \Bbb Z^r$. There's the projection map $\pi_2:G\to\Bbb Z^r$. Then $\pi_2\circ\phi:\Bbb Z^r\to\Bbb Z^r$ is onto. Thinking about matrices and determinants, this implies that $\pi_2\circ\phi$ is one-to-one. If $T$ is nonzero, then there is $a\in \Bbb Z^r$ such that $\phi(a)\ne0$ but $m\phi(a)=0$ for $m>1$. Then $\phi(ma)=0$, but this is a contradiction: $a\ne0$ in $\Bbb Z^r$ so $ma\ne0$, but $\phi$ is injective.

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  • $\begingroup$ May I know what do you mean by "Thinking about matrices and determinants, this implies that $\pi_2\circ\phi$ is one-to-one"? $\endgroup$ – Alan Wang Aug 3 '18 at 7:57
  • $\begingroup$ @AlanWang A homomorphism on $\Bbb Z^r$ is represented by a matrix with integer entries: this matrix must have full rank.... $\endgroup$ – Lord Shark the Unknown Aug 3 '18 at 8:01
  • $\begingroup$ Just to make sure that I think correctly, do you mean that we view a homomorphism on $\Bbb{Z}^r$ as a linear transformation and apply Rank-Nullity Theorem here? $\endgroup$ – Alan Wang Aug 3 '18 at 8:03
  • $\begingroup$ @AlanWang That's basically it: just employ some elementary linear algebra. $\endgroup$ – Lord Shark the Unknown Aug 3 '18 at 8:04

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