Let \(G\) be a finitely generated subgroup of \(\mathbb{Q}\) with generating set \(\frac{p_1}{q_1}\), …, \(\frac{p_n}{q_n}\) where \(p_i,q_i\) are integers. Setting \(q=q_1\cdots q_n\) we may write \(\frac{a_i}{q}=\frac{p_i}{q_i}\) for integers \(a_i\). Recall that \(\langle a_1,\ldots,a_n \rangle = \langle g \rangle\) as subgroups of \(\mathbb{Z}\) where \(g=\gcd(a_1,\ldots,a_n)\). Thus, every element of the form \(x=x_1a_1+\cdots+x_na_n\) for integers \(x_i\) can be written as \(x=cg\) for some integer \(c\). If \(y \in G\), then for some integers \(x_i\) we have \[y=x_1\frac{a_1}{q}+\cdots+x_n\frac{a_n}{q}=\frac{x_1a_1+\cdots+x_na_n}{q},\] so \(y=c\frac{g}{q}\) for some integer \(c\). Thus \(G = \langle\frac{g}{q}\rangle\) is cyclic. (Note that we cannot prove that \(G=\langle \frac{1}{r} \rangle\) for some integer \(r\).)

If \(\mathbb{Q}\) were finitely generated, then we would have \(\mathbb{Q}=\langle \frac{g}{q} \rangle\) as above. But \(\frac{g}{2q} \in \mathbb{Q} \setminus G\) gives a contradiction.

1. Given \(q + \mathbb{Z}\) in \(\mathbb{Q}/\mathbb{Z}\), consider the fractional part of \(q\) given by \(\{q\} = q -\lfloor q \rfloor\). Note that \(0 \le \{q\} < 1\) and \(q + \mathbb{Z} = \{q\} + \mathbb{Z}\). Thus we have the desired representative. If \(q,q'\) are two such respresentatives, then \(q-q'\) is an integer with \(|q-q'|<1\). Thus \(q-q'=0\) and so \(q=q'\).

2. Suppose \(x=\frac{a}{b} + \mathbb{Z} \in \mathbb{Q}/\mathbb{Z}\). We have a positive integer \(n=|b|\) such that \(n\frac{a}{b} \in \mathbb{Z}\). Thus \(nx = n\frac{a}{b} + \mathbb{Z} = \mathbb{Z}\) is trivial. Thus \(x\) has finite order \(n\). We observe that \(\frac{1}{n}\) has order \(n\) for any positive integer \(n\), so there are elements of arbitrarily large order.

3. We have already seen that \(\mathbb{Q}/\mathbb{Z}\) is torsion from part (b). Suppose \(x + \mathbb{Z}\) is in \(\mathbb{R}/\mathbb{Z}\) and has finite order \(n\). Then \(nx + \mathbb{Z}=\mathbb{Z}\) so \(nx \in \mathbb{Z}\). Since \(m=nx\) is an integer, \(x=m/n \in \mathbb{Q}\). Thus \(\mathbb{Q}/\mathbb{Z}\) is the torsion subgroup.

4. Consider the function \(\varphi : \mathbb{Q} \to \mathbb{C}^\times\) given by \(\varphi(z)=e^{2\pi i z}\). We check that \(\varphi(z+z')=e^{2\pi i (z+z')}=e^{2\pi i z}e^{2\pi i z'}=\varphi(z)\varphi(z'),\) so \(\varphi\) is a group homomorphism. Recall that \(e^{x}=1\) if and only if \(x=2\pi i k\) for an integer \(k\). Thus \(\ker(\phi)=\mathbb{Z}\). Thus \(\operatorname{im}(\varphi) \cong \mathbb{Q}/\mathbb{Z}\) by the first isomorphism theorem. It remains to prove that \(\operatorname{im}(\varphi)\) is the set of roots of unity. Since \(\mathbb{Q}/\mathbb{Z}\) is torsion, \(w^n=1\) for some integer \(n\) for all \(w \in \operatorname{im}(\varphi)\). Conversely, if \(w^n=1\), then \(1=|w^n|=|w|^n\) and \[0 \equiv \arg(z^n) \equiv n\arg(z) \pmod{2\pi}.\] Thus \(w = e^{2\pi i \frac{a}{n}}\) for some integer \(a\). Thus \(w \in \operatorname{im}(\varphi)\).

(We do not have to prove that \(N\) is a subgroup since it is a subgroup by definition. Note that the set \(\{ x^{-1}y^{-1}xy \mid x,y \in G \}\) is not necessarily a subgroup!)

Let \(g \in G\). Let \(\varphi_g(h)=ghg^{-1}\) be the conjugation map. Recall (or easily check) that \(\varphi_g\) is a group homomorphism. Thus \[\varphi_g(x^{-1}y^{-1}xy) = \varphi_g(x)^{-1}\varphi_g(y)^{-1}\varphi_g(x)\varphi_g(y)\] for all \(x,y \in G\). We have proved that if \(z\) is a generator of \(N\), then \(\varphi_g(z)\) is in \(N\). Since every element of \(N\) is a product of generators and their inverses, and \(\varphi_g\) is a group homomorphism, we conclude that \(\varphi_g(n) \in N\) for all \(n \in N\). Thus \(gNg^{-1} \subseteq N\) for all \(g \in G\) and we conclude \(N\) is normal.

To see that \(G/N\) is abelian, observe that \[ yxN=yxx^{-1}y^{-1}xyN=xyN\] for all \(x,y \in G\).