The Prüfer rank of an abelian group. 

If $p$ is a prime and $G$ an abelian group, the $p$-rank of $G$, $$r_p(G)$$ is defined as the cardinality of a maximal independent subset of elements of $p$-power order. Similarly, the $0$-rank or torsion-free rank $$r_0(G)$$ is the cardinality of a maximal independent subset of elements of infinite order. Also important is the Prüfer rank, often just called the rank of $G$, $$r(G)=r_0(G)+\max_p r_p(G).$$

Hi: Let's call the second term in the right side of the equality $L$. What guaranties the existence of $L$? I can have an infinite sequence of primes $p_1, p_2, ...$ with $r_{p_i}(G) \lt r_{p_{i+1}}(G)$. And there will clearly be no maximum. Is there a mistake in this text?
 A: You are correct that $\max$ is not technically correct, it should be $\sup$.
Now, Robinson states that the purpose of that section is to describe the structure of finite abelian groups, abelian groups with the maximal condition, and abelian groups with the minimal condition.
For finite abelian groups, $r_0=0$, each $r_p$ is finite, and $r_p(G)=0$ for almost all $p$; they are all bounded by $|G|$, and so $\max$ actually makes sense. 
The "maximal condition", as given by Robinson, is that every nonempty collection of subgroups has maximal elements; equivalently ACC on subgroups. For abelian groups, this is equivalent to being finitely generated (Proposition 4.2.8 in Robinson); for such a group, the torsion subgroup is finite, so again each $r_p$ is finite and it equals $0$ for almost all $p$, and $r_0$ is finite, so again $\max$ makes sense.
The "minimal condition" is that every nonempty collection of subgroups has  minimal elements, or equivalently that the group has DCC on subgroups. In Proposition 4.2.11 Robinson gives the Theorem of Kuros that says that an abelian group satisfies the minimal condition if and only if it is a direct sum of finitely many quasicyclic and cyclic groups of prime power order. For such a group, $r_0=0$, $r_p=0$ for almost all $p$, and all $r_p$ are finite, so again we get that the right hand side makes sense and $\max$ is adequate.
So for the context that Robinson is interested in, the expression for the Prüfer rank is sensible with $\max$.
That said, if we go to a more general setting, we can still make the definitions given if we use cardinals rather than finite quantities, and we replace $\max$ with $\sup$. In that case, since both $r_0$ and each $r_p$ is bounded above by $|G|$, the collection of cardinals is a set that is bounded above and so have a supremum, and the Prüfer rank is certainly at most $|G|$ (and could be equal to $|G|$; for example, the direct sum of $\aleph_0$ copies of $\mathbb{Z}$ has torsionfree rank $\aleph_0=|G|$. 
