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In other words, are there words $w_1$ and $w_2$ in the free group $F_2$ such that $F_2=\langle w_1, w_2\rangle$ but which are not a free basis for $F_2$?

I'm sure I'm missing a simple argument or a simple example.

user1729
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Sean Eberhard
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    If every surjection is an injection then the group is called Hopfian. It is an astonishing result of Mal'cev that every residually finite group is Hopfian. It is not difficult to show that $F_2$ is residually finite, so perhaps this would be a neat way of going. A proof of Mal'cev's result can be found in Robinson's book A course in the theory of groups (p165). It is only a paragraph long. – user1729 Apr 10 '13 at 09:41
  • (In the above comment, Mal'cev's result required that the group also be finitely generated: A finitely generated, residually finite group is Hopfian. Clearly $F_2$ is finitely generated. More information can be found about Hopficity on math.stackexchange here and here (and other places too.)) – user1729 Apr 10 '13 at 09:51
  • @user1729 Thanks, interesting result! – Sean Eberhard Apr 10 '13 at 09:52
  • The argument can also be found on wikipedia: http://groupprops.subwiki.org/wiki/Finitely_generated_and_free_implies_Hopfian – Seirios Apr 10 '13 at 10:20

2 Answers2

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In fact, it is true even for (finitely generated) residually free groups as a consequence of the following lemma:

Lemma: Let $G_1 \overset{\varphi_1}{\twoheadrightarrow} G_2 \overset{\varphi_2}{\twoheadrightarrow} G_3 \overset{\varphi_2}{\twoheadrightarrow} \dots$ be a sequence of epimorphisms between finitely generated residually free groups. Then all but finitely many epimorphisms are isomorphisms.

Indeed, if $\varphi : G \to G$ is an epimorphism but not an isomorphism (where $G$ is finitely generated and residually free), then the sequence $G \overset{\varphi}{\twoheadrightarrow} G \overset{\varphi}{\twoheadrightarrow} G \overset{\varphi}{\twoheadrightarrow} \dots$ contradicts the lemma.

Proof of the lemma: Let $S_1=(s_1^{(1)},...,s_n^{(1)})$ be a generator set of $G_1$; then $S_k:=\varphi_k(S_1)=(s_1^{(k)},...,s_n^{(k)})$ is a generator set of $G_k$ by surjectivity of $\varphi_k$. Let $V_k$ the set of $(M_1,...,M_n) \in SL_2(\mathbb{C})^n$ such that $r(M_1,...,M_n)=\operatorname{Id}$ for every relation $r$ of $G_k$.

Notice that $V_k$ is an affine algebraic variety in $\mathbb{C}^{4n}$ and $V_1 \supset V_2 \supset \dots$, so the sequence $(V_k)$ is eventually constant. To conclude, it is sufficient to show that if $\varphi_k : G_k \to G_{k+1}$ is not an isomorphism, then $V_{k+1} \subsetneq V_k$.

Let $r$ be a word such that $r$ over $S_{k+1}$ is a relation of $G_{k+1}$ but $r$ over $S_k$ is not a relation of $G_k$. $G_k$ being residually free, there exists a morphism $\rho : G_{k} \to \mathbb{F}_2$ such that $\rho(r) \neq 1$. Because $\mathbb{F}_2$ is a subgroup of $SL_2(\mathbb{C})$, you can suppose that $\rho : G_{k} \to SL_2(\mathbb{C})$.

Let $N_i= \rho(s_i^{(k)}) \in SL_2(\mathbb{C})$. Then $r(N_1,...,N_n)=\rho(r) \neq 1$ so $(N_1,...,N_n) \notin V_{k+1}$. If $w$ is a word over $S_k$ such that $w$ is a relation of $G_k$, then $w(N_1,...,N_n)=\rho(w)=\rho(1)=1$ so $(N_1,...,N_n) \in V_k$. $\square$

Seirios
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Here, for reference, is the proof of Mal'cev's result that every fintely generated residually finite group $G$ is Hopfian, or in other words that the only surjections $G\to G$ are the bijections.

Let $\phi:G\to G$ be a surjection, and suppose that $\phi(g)=e$ for some $g\neq e$. By surjectivity of $\phi$ we can find succesive preimages $(g_k)_{k\geq 1}$ such that $g_1 = g$ and $\phi(g_{k+1})=g_k$ for all $k\geq 1$. Let $\alpha:G\to K$ be a finite quotient such that $\alpha(g)$ is nontrivial. I claim the maps $\alpha\circ\phi^m:G\to K$ with $m\geq0$ are all distinct: indeed $\alpha\circ\phi^m$ and $\alpha\circ\phi^n$ ($n>m$) take different values on $g_{m+1}$. But if $G$ is finitely generated then it has only finitely many distinct homomorphisms $G\to K$, a contradiction.

Sean Eberhard
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