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.


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.

  • $\begingroup$ 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.... $\endgroup$ – Angina Seng 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. $\endgroup$ – Angina Seng Aug 3 '18 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.