What is the difference between a free Abelian group (of say a finite basis) and a finitely generated Abelian group? Is the difference that the generating set of the finitely generating Abelian group need not be linearly independent? I understand that both the basis and the finite generating set span the Abelian group. Thanks for clarifying.
 A: Every group $G$ is the quotient of a free group $F$. To say that $G$ is finitely generated means that $F$ has a finite set of generators. Saying basis and linearly independent is the right idea, but isn't quite proper, since we reserve those words for vector spaces. A finitely generated group has a finite generating set and possibly some relations among those generators.

To address your comment and expand a bit, every group $G$ has a presentation as a set of generators and relations. For example, this abelian group:
$$G = \left\langle a, b \mid aba^{-1}b^{-1}, a^3, b^7 \right\rangle$$ 
Now this group presentation is really shorthand for $G$ being the quotient of the free group $\langle a,b\rangle$ by the normal subgroup generated by the elements $\langle aba^{-1}b^{-1}, a^3, b^7 \rangle$. If you want to get fancy and use more categorical language, that presentation of $G$ will be the cokernel of the map $\langle x,y,z \rangle \to \langle a, b \rangle$, such that $x \mapsto aba^{-1}b^{-1}$, $y \mapsto a^3$, and  $z \mapsto b^7$. 
A: I personally find the following perspective useful. This also generalizes nicely to other situations. 
An abelian group $G$ is free whenever there is an isomorphism 
$$
\alpha\colon\mathbf{Z}^I \to G
$$
for some index set $I$.
The usual generators of $\mathbf{Z}^I$ are then mapped to generators of $G$ by $\alpha$. 
An abelian group $H$ is finitely generated whenever there is a surjective homomorphism
$$
\beta\colon\mathbf{Z}^n \to H. 
$$
for some natural number $n\in\mathbf{N}$. 
Again, $\beta$ maps generators to generators, but now they may be subject to relations. In other words, there is some subgroup $R$ of $\mathbf{Z}^n$ such that $H$ is isomorphic to $\mathbf{Z} ^n / R$.
In the situations above, if $I$ happens to be a finite set, or if $\beta$ happens to be injective (or equivalently, $R=0$), then $G$ or $H$ respectively are both free and finitely generated. 
