In the subject "Algebraic Topology" we define the Betti's number as the greater number $\beta_p$ such that a family $\{z^i_p\}_{i=1}^{\beta_p}$ of $p-$cicles are linearly independent (i.e. there's no exists a family $\{\lambda_i\}_i\subset \mathbb{Z}$ such that $\sum_{i} \lambda_i z^i_p$ is homologous to $0$).
How can I see that this definition of Betti's number is equivalent to be the rank of the free part of the $p-$homology group?
Thank you in advance.
Each member $z_p^i$ of this family of $p$-circles in the space $X$ defines a $p$-cycle in $X$, and the condition that the $z_p^i$'s be linearly independent (in the sense described in your question) imply that they are each a free generator in the p-th homology. In particular, since $\beta_p$ is the greatest number of such linearly independent $p$-circles, the image of $\{ z_p^i \}_{i=1}^{\beta_p}$ in homology generates the maximal free subgroup of $H_p(X, \mathbb{Z})$ (i.e. the free part of the homology group). Therefore, $\beta_p$ is precisely the rank of the free part of $H_p(X,\mathbb{Z})$.
Abstract away from the topology. Show that for any finitely generated abelian group $A$, the rank of the free part of $A$ is the maximal size of a linearly independent subset. The reason we can ignore torsion is that even a single torsion element $a\in A$ is not linearly independent, since exactly the definition of torsion is that $na=0$ for some nonzero $n$.
