Definition of the rank of an abelian group According to wikipedia, the rank of an abelian group $G$ is the size of the largest free abelian group contained in $G$. So, is the rank of $G$ equal to:
$$\sup \{|F|\mid {F\text{ is a subgroup of }G\text{, }F\text{ is a free abelian group} }\}$$
?
here, $|F|$ is the cardinality of $F$.
I'm looking for a symbolic definition in general case. What is meant by largest in wikipedia?
 A: That's not what wikipedia says. It says the rank is the cardinality of a maximal linearly independent subset. Here, maximal means not properly contained in another linearly independent subset. Linearly independent is defined in the body of the entry: using additive notation, $\{x_1,\ldots,x_n\}$ is a linearly independent set if
$$
a_1x_1+\cdots+a_nx_n=0 \Rightarrow \forall i\;a_i=0
$$
where the $a_i\in\mathbb{Z}$ and the $x_i$ are elements of the group. An infinite subset is linearly independent if all of its finite subsets are. The wikipedia entry gives a reference to Lang's Algebra, but unfortunately Lang actually gives a different (though equivalent) definition of rank.
The main complicating factor, at least in the finitely generated case, are torsion elements, i.e., elements of finite order. However, any linearly independent set cannot contain any torsion elements, since $ax=0\Rightarrow a=0$ follows from linear independence, and this means that $x$ is not a torsion element.
Example: $A=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_2$. The set $\{(1,0,0),(0,1,0)\}$ is a maximal linearly independent set. There are others, for example $\{(2,0,0),(0,3,0)\}$, but they all have the same cardinality. (This is not completely trivial to prove, though it's not super-hard either, once you have enough background.) 
The second sentence of the wikipedia entry says, "The rank of $A$ determines the size of the largest free abelian group contained in $A$."  This is not a definition. However, it is inspired by one:
$$
\text{rank }A = \max \{\dim F | F\text{ a free abelian subgroup of }A\}
$$
Of course we then need a definition for the dimension (aka rank) of a free abelian group. One definition: since the tensor product $\mathbb{Q}\otimes_\mathbb{Z} F$ is a vector space over $\mathbb{Q}$, just define $\dim F$ to be the vector space dimension of $\mathbb{Q}\otimes_\mathbb{Z} F$.
Since tensoring with $\mathbb{Q}$ kills the torsion subgroup, the rank of $A$ is just $\dim \mathbb{Q}\otimes_\mathbb{Z} A$; no need to look at the free abelian subgroups.
