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Let $:F_n$ denote the free group of rank $n$. How can I solve the exercise 7.6.3.(b), page 234, in Massey's Algebraic Topology?

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I'm guessing there's been made a mistake and (b) actually reads "If $G$ is a free group of finite rank and $N$ is a nontrivial normal subgroup of $G$, then $G/N$ is not isomorphic to $G$", since $N=\{1\}$ gives a counterexample.

So, how can I prove:

$$\{1\}\neq N\trianglelefteq F_n\Longrightarrow F_n\ncong F_n/N?$$

I don't know how to use 7.6.3:

If $f:F_n\rightarrow F_n$ is a surjective homomorphism, it is an isomorphism.

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did you prove the first part of the statement yet (what precedes the questions $(a)$ and $(b)$)? –  Olivier Bégassat Aug 18 '11 at 9:48
    
@Olivier: yes, just use the Grushko theorem: If $\varphi\!:F_X\!\rightarrow\!\amalg_{i\in I}G_i$ is a surjective homomorphism, then there exists a decomposition $F_X=\amalg_{i\in I}F_{X_i}$ such that $\forall i\!\in\!I\!:\:\varphi(F_{X_i})=G_i$. But how can I use that to prove (b)? –  Leon Aug 18 '11 at 9:53

2 Answers 2

up vote 3 down vote accepted

The first part of your question, that if $f: F_{\{a_1, \ldots, a_n\}}\rightarrow F_{\{b_1, \ldots, b_n\}}$ is an epimorphism then it is an isomorphism is a property called Hopfian. It is a very nice property, and I, personally, refuse to believe that any group is non-Hopfian (I know that Baumslag-Solitar groups are, as in $\mathbb{Z}^{\infty}$, but that doesn't mean I believe they are...).

Now, the part (b) is just a basic application of the first isomorphism theorem ($G/\ker\phi\cong \operatorname{im}\phi$). If $F_n\cong F_n/N$ then there exists some epimorphism, $\phi: F_n\rightarrow F_n$, which has kernel $N$ (why?). As $F_n$ is Hopfian, $N$ must be trivial, and your done.

Really, the tricky bit is proving Hopfian. If you can prove that $F_n$ is residually finite, then you're doing even better (residually finite is a strictly stronger property, which free groups also have - $G$ is residually finite if for every non-trivial element $w$ there exists a finite group $H_w$ such that there exists $\phi: G\rightarrow H_w$ an epimorphism, and $w\phi\neq 1$). The neatest proof of the residual finiteness of free groups that I know of goes as follows:

  1. If $G$ is finitely generated and residually finite then $\operatorname{Aut}(G)$ is residually finite.
  2. As $\operatorname{GL}_2(\mathbb{Z})\cong \operatorname{Aut}(\mathbb{Z}\times\mathbb{Z})$, we have that $\operatorname{GL}_2(\mathbb{Z})$ is residually finite.
  3. $F(a, b)\leq \operatorname{GL}_2(\mathbb{Z})$, by the following embedding. $$\begin{align*}a&\mapsto\left(\begin{array}{cc}1&2\\0&1\end{array}\right)\\b&\mapsto\left(\begin{array}{cc}1&0\\2&1\end{array}\right)\end{align*}$$
  4. As subgroups of residually finite groups are residually finite, $F(a, b)$ is residually finite.
  5. Every countable free group embeds into $F(a, b)$, hence every countable free group is residually finite.

Every step here is non-trivial. However, steps (2)-(5) are all well-documented and probably covered in whatever course you are studying (I think John Meier's book Graphs, groups and trees is probably the best reference for these, especially step (3) where he uses the ping-pong lemma). So I need to tell you about step (1). However, step (1) is why this is so wonderful and beautiful and amazing. Step (1) appeared in a 1-page (not including references) paper of G. Baumslag. His proof is cunning and simple.

Theorem[G. Baumslag] Suppose $G$ is finitely generated and residually finite. Then $G$ has residually finite automorphism group.

The proof uses the following definition: a subgroup $N$ of a group $G$ is called \emph{characteristic} if $N\phi=N$ for all $\phi\in\operatorname{Aut}(G)$.

Begin by noting that if $H\leq_fG$ then there exists a characteristic subgroup $N$ of $G$ with $N\leq_fH$, which is obtained by intersecting the subgroups of index $|G:H|=:n$. The resulting group is characteristic because automorphisms preserve index, and it has finite index because there are only finitely many subgroups of index $n$ as $G$ is finitely generated.

Now, suppose $\phi\in\operatorname{Aut}(G)$ is non-trivial. We shall find a finite group $K$ such that $\operatorname{Aut}(G)\rightarrow\operatorname{Aut}(K)$ and $\phi$ is not mapped to the identity under this homomorphism. To do this, note that as $\phi$ is non-trivial there exists some $g\in G$ such that $g^{-1}(g\phi)\neq 1$. Choose $x\neq g^{-1}(g\phi)$. Then there exists some characteristic subgroup $N_x\unlhd_f G$ with $x\not\in N_x$. Set $K=G/N_x$. Then $\phi$ acts on $G/N_x$ because $N_x$ is characteristic, and as $g^{-1}(g\phi)\not\in N_x$ we have that $gN_x\neq (gN_x)\phi$. Therefore, the action of $\phi$ on $K$ is non-trivial, as required.

Note: An alternative proof that $\operatorname{GL}_2(\mathbb{Z})$ is residually finite, so an alternative to steps (1) and (2) combined, is given here.

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From wiki: A group $G$ Hopfian if every epimorphism $G \rightarrow G$ is an isomorphism. Equivalently, a group is Hopfian iff it is not isomorphic to any of its proper quotients. A group G is co-Hopfian if every monomorphism $G \rightarrow G$ is an isomorphism. Equivalently, $G$ is not isomorphic to any of its proper subgroups. Hmm very interesting, thanks. –  Leon Aug 18 '11 at 10:23
    
Could you please explain how I should use the first isomorphism theorem to find and epimorphism $F_n\rightarrow F_n$ with kernel $N$? By "first isomorphism theorem", do you actually mean "Factorization Theorem" (every $\varphi:G\rightarrow H$ with $N\leq ker\varphi$ factors uniquely through $G/N$) or "Homomorphism Theorem" ($A/ker\varphi\cong im\varphi$) or really the "first isomorphism theorem" ($A/B\cong\!(A/C)/(B/C)$)? –  Leon Aug 18 '11 at 10:58
    
I'd guess you just precompose with the quotient projection. –  Miha Habič Aug 18 '11 at 16:47
    
Ah, if $\pi:F_n\rightarrow F_n/N$ is the quotient projection and $f:F/N\rightarrow F_n$ is the isomorphism, then $f\circ \pi$ is the desired homomorphism $F_n\rightarrow F_n$ with kernel $N$. Don't really know what the isomorphism theorems have to do with it. Thanks @Miha. –  Leon Aug 19 '11 at 9:25
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@Leon: Free groups of infinite rank are not hopfian: pick a proper subset $B$ of the set of original generators $A$ of the same cardinality. Define a map by sending $A\setminus B$ to 1 and $B$ to $A$ (bijectively) and extend this to a noninjective epimorphism. –  Miha Habič Aug 19 '11 at 19:30

Let $N$ be a normal subgroup of $F = F_{n}$. The mapping from $F$ upon the quotient group $F/N$ which takes $f$ in $F$ to the coset $fN$ is an epimorphism; call it $\eta$. Suppose there is an isomorphism $\varphi$ from $F/N$ to $F$. Composing these two maps, we get an epimorphism $\varphi\circ\eta : F\to F/N\to F$ of $F$ upon itself. Now use the first part to conclude that it is an isomorphism. In particular, the isomorphism $\varphi\circ\eta$ is injective, so $\eta$ is injective. Therefore, the kernel $N$ of $\eta$ is trivial.

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thank you for a clarification –  Leon Aug 19 '11 at 9:31

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