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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?

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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 $m=n$.

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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).

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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?

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    $\begingroup$ We can provide the following counterexample to the conjecture that $X_1$ and $X_2$ must be isomorphic in the last sentence. Take $G$ to be the free functor $\mathbf{Set}→\mathbf{Vect}_ℝ$ and $F$ the forgetful functor $\mathbf{Vect}_ℝ→\mathbf{Set}$. Then $\newcommand{\H}{\mathrm{Hom}}\H(G(X),ℝ^2)≅\H(G(X),ℝ)$ but $ℝ^2 ≇ ℝ$. I add that this is not the Yoneda lemma that is used here, but only the fact that if $X≅Y$, then $\H(X,\bullet)≅\H(Y,\bullet)$. $\endgroup$
    – Idéophage
    Sep 8 '18 at 22:12
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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.

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  • $\begingroup$ I've only had about a month's study of abstract algebra; and am really interested in the topic but need a bit of a more "basic" answer to be able to understand it; if its a question that can be attacked in a more elementary way. Thanks a ton though! $\endgroup$
    – Steve
    Dec 19 '10 at 5:47
  • $\begingroup$ @Steve Feel free to ask about specific words or concepts that you don't know. Also, the question itself has been answered by Noah. I was just putting it into a more general context. $\endgroup$
    – Alex B.
    Dec 19 '10 at 6:01

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