Short Exact Sequences & Rank Nullity

If you're not familiar with tensor products, then learning them for this problem is overkill. But if you are familiar with them, then I think the following solution is nice. $\newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}}$

For a finitely generated abelian group $A$, note that $\mathrm{rank}(A) = \dim_\Q(A\otimes_\Z \Q)$. One can show (basically following Arturo's answer) that given a short exact sequence of abelian groups $0\to A\to B\to C\to 0$, the sequence obtained by tensoring with $\Q$ is also exact: $$0\to A\otimes_\Z \Q \to B\otimes_\Z \Q \to C\otimes_\Z \Q \to 0$$

Now the statement about ranks boils down to an easy statement about how dimensions of vector spaces add in a short exact sequence.

Pick a maximal linearly independent subset $\{c_{\beta}\}$ of $C$. Now push the $a_{\alpha}$ to $B$ using $f$, and for each $c_{\beta}$ pick $c'_{\beta}\in B$ such that $g(c'_{\beta}) = c_{\beta}$.

Now suppose that you have a finite linear combination of the $a_{\alpha}$ and the $c'_{\beta}$ that is equal to $0$, $$n_{\alpha_1}f(a_{\alpha_1}) + \cdots + n_{\alpha_k}f(a_{\alpha_k}) + m_{\beta_1}c'_{\beta_1} + \cdots + m_{\beta_{\ell}}c'_{\beta_{\ell}} = 0.$$ Use $g$ to get a conclusion about the $m_{\beta_j}$; then use $f$ to get a conclusion about the $n_{\alpha_i}$. This will give you that $\mathrm{rank}(B)\geq \mathrm{rank}(A)+\mathrm{rank}(C)$.

Do you also need help with the converse inequality?