I get a contradiction in the theory of free abelian groups. What am I doing wrong? Hi: The definition I'll use is this: Let $F$ be an abelian group and $X$ a subset of $F$. Then $F$ is a free abelian group on $X$ if for every abelian group $G$ and every function $f$ from $X$ to $G$ there is a homomorphism $\phi$ from $F$ to $G$ that extends $f$.
Let $G$ be a finite group and $X$ a subset of $G$. Let $F$ be the free abelian group on $X$. Then $F=\langle X\rangle$ and so $F\subseteq G$. That is, every finite group has an infinite subgroup. What am I doing wrong?
EDIT: It will be easier to make myself clear working with free groups. I'll quote from Derek Robinson, A Course in the Theory of Groups, 2nd ed.

From this a free group is not only always free on a subset but additionally that subset generates it. If $G$ is a group and $X$ is a subset, however, indeed there will exist a free group on $X$ but I am unable to show it will be generated by $X$ based in the above quote. Which is very natural, of course. Thanks for the posts. Honestly none of the feedback, up to now,  throws light in the paradox (paradox for me, of course).
 A: You are using $\langle X \rangle$ to mean two different things, and conflating them: 


*

*You are using it to mean the free abelian group on $X$. 

*You are using it to mean the subgroup of $G$ generated by $X$. 


These are not the same thing, but you assume that they are. 
A: The $X$ in the free abelian group definition is a collection of formal symbols. They are assumed to have the minimum properties necessary for $\langle X\rangle$ to be an abelian group. That is what it means to be "free" in this context.
However, for any $\Xi\subseteq G$ for a finite group $G$, there are extra constraints on the subgroup $\langle \Xi\rangle_G$ of $G$; for instance, each element of $\langle \Xi\rangle_G $ is of finite order. Thus it is not free.
A: Consider $G=\mathbb{Z}/5\mathbb{Z}$ and $[1]\in G$, let $X=\{[1]\}$ $F=<X>=G$, let $f:X\rightarrow\mathbb{Z}$ defined by $f([1])=1$, $f$ cannot be extended to $G$ in $g$, since we must have $g(5.[1])=5.f([1])=5$ and $g(5.[1])=g([0])=0$ since $g$ is a morphism of groups. 
