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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 Lampret Aug 18 '11 at 9:53

2 Answers

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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 ($G/ker\phi\cong im\phi$). 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 $w\phi\neq 1$).

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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 Lampret 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 Lampret 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 Lampret 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
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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 Lampret Aug 19 '11 at 9:31

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