If $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is a short exact sequence of abelian groups, then rank $B$ = rank $A$ + rank $C$ 
Prove that if  $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is a short exact sequence of abelian groups, then rank $B$ = rank $A$ + rank $C$. (Hint: Extend a maximal independent subset of $A$ to a maximal independent subset of $B$)

I'm not sure what it means to extend a maximal indepdent subset of $A$ to a maximal independent subset of $B$.  I see that since it's a short exact sequence, $A \approx$ im $A$, $B/$ im $ A \approx C$, but from here I can't see any relationship.
An abelian group $G$ has rank $r$ iff there exists a free abelian subgroup $F$ with rank $F = r$ (cardinality of the basis) and $G/F$ is torsion.
 A: Hint:
Tensor the exact  sequence by $\mathbf Q$. The rational rank of an abelian group is the dimension of the resulting $\mathbf Q$-vector space.
A: Call the maps $A \to B$ and $B \to C$, $f$ and $g$ respectively.
Let $a_1,\dots, a_m$ be a maximal independent set in $A$ and $c_1,\dots,c_n$ a maximal independent set in $C$. Define $b_1 = f(a_1), \dots, b_m = f(a_m)$ and choose $b_1', \dots, b_n'$ such that $g(b_1') = c_1, \dots, g(b_n') = c_n$.
I'm going to use some spoilers here because the proofs are short and you should try them on your own.
Step 1: $b_1, \dots, b_m$ are linearly independent

 Suppose $\lambda_1 b_1 + \dots + \lambda_m b_m = 0$. Then $f(\lambda_1 a_1 + \dots + \lambda_m a_m) = 0$ and since $f$ is injective, $\lambda_1 a_1 + \dots + \lambda_m a_m = 0$. But $a_1,\dots, a_m$ are linearly independent so $\lambda_1 = \dots = \lambda_m = 0$.

Step 2: $b_1', \dots, b_n'$ are linearly independent

 Suppose $\mu_1 b_1' + \dots + \mu_n b_n' = 0$. Apply $g$ to get $\mu_1 c_1 + \dots + \mu_n c_n = 0$. Since $c_1,\dots, c_n$ are linearly independent, $\mu_1 = \dots = \mu_n = 0$.

Step 3: $b_1,\dots,b_m, b_1', \dots, b_n'$ are linearly independent (essentially the same as steps 1 and 2)

 Suppose $\lambda_1 b_1 + \dots + \lambda_m b_n + \mu_1 b_1' + \dots + \mu_n b_n' = 0$. Apply $g$ to get $\mu_1 c_1 + \dots + \mu_n c_n = 0$. Hence $\mu_1 = \dots = \mu_n = 0$ and then $\lambda_1 b_1 + \dots + \lambda_m b_n = 0$ implies $\lambda_1 = \dots = \lambda_m = 0$.

Step 4: $b_1,\dots,b_m, b_1', \dots, b_n'$ is maximal
Suppose not. Let $d' \in B$ be added to this list to get a larger independent list. Now $c_1,\dots,c_n, g(d')$ is not independent because $c_1,\dots,c_n$ is maximal. Thus $\nu' g(d') = \mu_1 c_1 + \dots + \mu_n c_n$ with $\nu' \ne 0$. Thus for some $d \in \ker g = \operatorname{im} f$ we have $\nu' d' = d + \mu_1 b_1' + \dots + \mu_n b_n'$. Let $a = f^{-1}(d) \in A$. Then $a, a_1,\dots, a_m$ is not linearly independent so $\nu a = \lambda_1 a_1 + \dots + \lambda_m a_m$ for some $\nu \ne 0$. Applying $f$, this says, $\nu d = \lambda_1 b_1 + \dots + \lambda_m b_m$. It follows that
$$ \nu\nu' d' = \lambda_1 b_1 + \dots + \lambda_m b_m +  \nu \left( \mu_1 b_1' + \dots + \mu_n b_n' \right) $$
where the coefficient $\nu\nu' \ne 0$. This is a contradiction.
