$\langle X|\emptyset\rangle\ncong\langle X|R\rangle$ for finitely presented groups (exercise in Massey) 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?

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.
 A: 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...). Some background on Hopfian groups can be found here.
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 that finitely generated free groups are Hopfian.
Now, a group $G$ 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 $\phi(w)\neq 1$. If $G$ is finitely generated and residually finite then $G$ is Hopfian. See here for a proof. Therefore, if we can prove that free groups are residually finite then we have shown that they are Hopfian. Whoop! The neatest proof of the residual finiteness of free groups that I know of goes as follows:


*

*If $G$ is finitely generated and residually finite then $\operatorname{Aut}(G)$ is residually finite.

*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.

*$F(a, b)\leq \operatorname{GL}_2(\mathbb{Z})$, by the following embedding. $$\begin{align*}a&\mapsto\left(\begin{array}{cc}1&2\\0&1\end{array}\right)\\b&\mapsto\left(\begin{array}{cc}1&0\\2&1\end{array}\right)\end{align*}$$

*As subgroups of residually finite groups are residually finite, $F(a, b)$ is residually finite.

*Every countable free group embeds into $F(a, b)$, hence every countable free group is residually finite.


Every step here is non-trivial, but not too hard. They are also all interesting. 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 $\phi(N)=N$ 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 $g^{-1}\phi(g)\neq 1$. Choose $x\neq g^{-1}\phi(g)$. 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 $g^{-1}\phi(g)\not\in N_x$ we have that $gN_x\neq \phi(gN_x)$. Therefore, the action of $\phi$ on $K$ is non-trivial, as required.
*This answer is basically an excuse to tell you about step (1) and the theorem of Baumslag. It is one of my all-time favourite proofs!
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.
A: 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.
