Generators of a free group If G is a free group generated by n elements, is it possible to find an isomorphism of G with a free group generated by n-1 (or any fewer number) of elements?
 A: Dave R's answer highlights a more general principle: by the Yoneda lemma, an object $X$ in a category is determined up to isomorphism by the behavior of the functor $F_X = \text{Hom}(X, -)$, so to show that two objects $X, Y$ are not isomorphic it suffices to show that the corresponding functors $F_X, F_Y$ are not isomorphic.  (In particular it suffices to show the existence of an object $Z$ such that $\text{Hom}(X, Z)$ has a different size from $\text{Hom}(Y, Z)$.)  
The free groups $F_n$ represent some special functors $\text{Grp} \to \text{Set}$: namely, $\text{Hom}(F_n, G)$ is precisely the set $G^n$ of $n$-tuples of elements of $G$.  This is a manifestation of the adjunction between the free group functor $\text{Set} \to \text{Grp}$ and the underlying set functor $\text{Grp} \to \text{Set}$, and by setting $G$ to any nontrivial finite group (Dave R's answer uses $\mathbb{Z}/2\mathbb{Z}$) it is not hard to see that these functors are all nonisomorphic. 
More generally, I believe one can say the following.  Let $F : D \to C$ and $G : C \to D$ be an adjunction, hence
$$\text{Hom}_C(FX, Y) \simeq \text{Hom}_D(X, GY)$$
and suppose that the functor $G$ is essentially surjective.  (In this case $G$ is the forgetful functor $\text{Grp} \to \text{Set}$ and $F$ is the free group functor $\text{Set} \to \text{Grp}$; then it is a classical result that $G$ is essentially surjective.)  Then by another application of the Yoneda lemma, I believe the functors $\text{Hom}_C(FX_1, -)$ and $\text{Hom}_C(FX_2, -)$ are isomorphic if and only if the objects $X_1, X_2$ are isomorphic in $D$.  Can anyone confirm this?
A: To add to Noah Snyder's answer: abelian groups are $\mathbb{Z}$-modules, and the ring $\mathbb{Z}$ has the Invariant Basis Number property. That means that for free $\mathbb{Z}$-modules, the rank is well-defined. More generally, all commutative rings have this property, as well as all left-Noetherian rings ($\mathbb{Z}$ is both). The wikipedia article gives an example of a ring that doesn't have this property, so modules over that ring can exhibit the strange behaviour that you are looking for.
A: Here is another approach. Let $G$ be a free group on $m$ generators, and let $H$ be a free group on $n$ generators. There are exactly $2^m$ homomorphisms from $G$ to a group of order two, since each generator can be mapped in two ways. Likewise, there are $2^n$ homomorphisms from $H$ to a group of order two.
If $G$ and $H$ are isomorphic, then they have the same number of homomorphisms to a group of order two. Therefore $2^m = 2^n$, which implies $m=n$.
A: No, if $F_n$ and $F_m$ are isomorphic, then their abelianizations are also isomorphic, which only happens when n=m (for example, reduce mod p and count points). 
