Let $G$ be a group generated by two elements $a$ and $b$. Suppose $G$ is a free group of rank 2. Is it true that $G=\langle a\rangle * \langle b\rangle$?

I think the problem is that the definitions I know of free group and free product are via universal properties, so I don't understand the equality in $G=\langle a\rangle * \langle b\rangle$. How should I interpret such equality, then?


(First I was trying to prove that $G$ is a freely generated by $a$ and $b$, but I end up getting that $G$ is generated by some image of $a$ and $b$ under an injective map. And from there I cannot get the desired equality.)

  • $\begingroup$ I think the equality just means: the map $\langle a\rangle * \langle b\rangle \to G$ induced by the inclusions (e.g. $\langle a\rangle \to G$) is an isomorphism. $\endgroup$ – Dylan Moreland Sep 5 '11 at 2:34
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    $\begingroup$ @b27: At first I thought your main difficulty was proving that the free group on $a$ and $b$ is equal/canonically isomorphic to the free product $\langle a\rangle*\langle b\rangle$. On re-reading, it seems that perhaps your main difficulty lies in proving that if $G$ is free of rank $2$ and is generated by two elements $a,b$, then in fact it is free on $\{a,b\}$. Could you clarify which one it is (or if it is both)? $\endgroup$ – Arturo Magidin Sep 5 '11 at 3:27

Yes, the free product of the free group of rank $n$ and the free group of rank $m$ is isomorphic to the free group of rank $n+m$. Note also that if $G$ is free of rank $2$, and two elements $a$ and $b$ generate $G$, then $a$ and $b$ must freely generate $G$. One way to see this is to invoke the fact that a free group of finite rank is Hopfian. Another is to assume there is a nontrivial reduced word in $a$ and $b$ equal to the identity, take a free generating set $x$ and $y$, express $a$ and $b$ in terms of $x$ and $y$, and replace them in the nontrivial word expressing the identity; this will yield a nontrivial word in $x$ and $y$ equal to the identity (some work needs to be done here, of course), contradicting the choice of $x$ and $y$ as a free generating set. And there are other ways, of course.

An easy way to get this is as a corollary to the fact that the free group functor is the left adjoint of the underlying set functor. That is, for every group $G$ and every set $X$, $$\mathcal{G}roup(\mathbf{F}(X),G) \longleftrightarrow \mathcal{S}et(X,\mathbf{U}(G)),$$ where $\mathbf{F}(X)$ is the free group on the set $X$ and $\mathbf{U}(G)$ is the underlying set of the group $G$.

Because the free group functor is a left adjoint, it sends coproducts to coproducts. That is, the coproduct of two free groups $F(X)$ and $F(Y)$ in the category of groups is the free group on the coproduct of $X$ and $Y$ in the category of sets. The coproduct in the category of groups is the free product, and the coproduct in sets is the disjoint union. Therefore, there is a natural isomorphism $$F(X\amalg Y) \cong F(X)*F(Y).$$

You can also prove it directly from the universal property: the universal property of the free group on $X\amalg Y$ (the disjoint union) is that for every set-theoretic map $f\colon X\amalg Y\to G$ to a group $G$, there is a unique group homomorphism from $F(X\amalg Y)\to G$ that extends $f$. On the other hand, the universal property of the free product $F(X)*F(Y)$ is that for every pair of group homomorphisms $\varphi\colon F(X)\to G$ and $\psi\colon F(Y)\to G$, there is a unique group homomorphism $\Psi\colon F(X)*F(Y)\to G$ such that $\Psi\circ i_{F(X)}=\varphi$ and $\Psi\circ i_{F(Y)}=\psi$, where $i_{F(X)}\colon F(X)\to F(X)*F(Y)$ and $i_{F(Y)}\colon F(Y)\to F(X)*F(Y)$ are the canonical inclusions.

A set-theoretic map $f\colon X\amalg Y\to G$ is equivalent to a pair of maps $g\colon X\to G$ and $h\colon Y\to G$; the map $g\colon X\to G$ induceds a map $\varphi\colon F(X)\to G$, while the map $h\colon Y\to G$ induces a map $\psi\colon F(Y)\to G$, which in turn induces a map $F(X)*F(Y)\to G$; it is now straightforward to verify that this map extends $f$, and that it is unique, so that $F(X)*F(Y)$ has the universal property of $F(X\amalg Y)$, and therefore they are isomorphic.

Or: the inclusions $X\to F(X)\to F(X)*F(Y)$ and $Y\to F(Y)\to F(X)*F(Y)$ induce an inclusion $X\amalg Y\to F(X)*F(Y)$, which in turn induces a morphism $F(X\amalg Y)\to F(X)*F(Y)$. Conversely, the map $X\to F(X\amalg Y)$ induces a map $F(X)\to F(X\amalg Y)$, and $Y\to F(X\amalg Y)$ induces a map $F(Y)\to F(X\amalg Y)$, and these two maps together induce a map $F(X)*F(Y)\to F(X\amalg Y)$. It is now easy to verify that the induces maps $F(X\amalg Y)\to F(X)*F(Y)$ and $F(X)*F(Y)\to F(X\amalg Y)$ are inverses of each, in the usual abstract nonsense argument about their compositions having the same universal property as the corresponding identity.

P.S. The equality $F(a,b)=\langle a\rangle*\langle b\rangle$ presumably just means that there is a unique isomorphism between the two objects that maps $\{a\}$ and $\{b\}$ to themselves as the identity; this is a consequence of their respective universal properties/adjointness of free group construction.

  • $\begingroup$ It doesn't appear to me that we're told that $G$ is free on $\{a, b\}$. That seems harder (but I have a very superficial knowledge of free groups). $\endgroup$ – Dylan Moreland Sep 5 '11 at 2:53
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    $\begingroup$ @Dylan: If it is generated by two elements, and free of rank 2, then it is free in those two elements; otherwise, you would have that the free group on two elements is a proper quotient of the free group on two elements, i.e., a proper quotient of itself. But free groups of finite rank are Hopfian, so they are not isomorphic to proper quotients of themselves. But I'll add that. $\endgroup$ – Arturo Magidin Sep 5 '11 at 3:00
  • $\begingroup$ @Arturo: Ok, $G$ is freely generated by $a$ and $b$. Could you please clarify how do we get the unique isomorphism from $G$ to $\langle a\rangle*\langle b\rangle$? $\endgroup$ – b27 Sep 5 '11 at 3:19
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    $\begingroup$ @b27: Uniqueness here refers to the only map that respects the set-theoretic inclusions $\{a,b\}\hookrightarrow G$, $\{a\}\hookrightarrow\langle a\rangle\hookrightarrow \langle a\rangle*\langle b\rangle$, and $\{b\}\hookrightarrow\langle b\rangle \hookrightarrow \langle a\rangle*\langle b\rangle$. $\endgroup$ – Arturo Magidin Sep 5 '11 at 3:26
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    $\begingroup$ @b27: Compose it with the map you get in the other direction from the universal properties: you have a map from $\langle a\rangle * \langle b\rangle$ to $G$ induced by the embeddings $\{a\}\hookrightarrow G$ and $\{b\}\hookrightarrow G$. Composing the map $G\to\langle a\rangle * \langle b\rangle \to G$ is an endomorphism of $G$ that maps $a$ to $a$ and $b$ to $b$, so by the universal property of free groups must be the identity. Composing them the other way and invoking the unversal property of $\langle a\rangle*\langle b\rangle$ gives that the other composition is also the identity. $\endgroup$ – Arturo Magidin Sep 5 '11 at 3:41

There is a general result: if $X$ is a set of $n$ elements of the free group $F$ on $n$ generators, and $X$ generates $F$. then $X$ generates $F$ freely. This result can be deduced from a powerful theorem known as Grusko's theorem, of which you will find one proof (in fact of a generalisation!) in the downloadable

Higgins, P.J. Notes on categories and groupoids, Mathematical Studies, Volume 32. Van Nostrand Reinhold Co. London (1971); Reprints in Theory and Applications of Categories, No. 7 (2005) pp 1--195.

I believe there is another proof of the first statement in the book by Lyndon and Schupp on Combinatorial Group Theory.


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