Let \(a,b\in G\text{.}\) Then

as desired.

We define \(\phibar: G/N\to G'\text{,}\) as indicated above, by \(\phibar(aN)=\phi(a)\) for every \(aN\in G/N\text{.}\) Since \(\phibar\) is defined using coset representatives, we must first show that \(\phibar\) is well-defined. So let \(aN=bN\in G/N\text{.}\) Then \(\phibar(aN)=\phi(a)\) and \(\phibar(bN)=\phi(b)\text{,}\) so we must show that \(\phi(a)=\phi(b)\text{.}\) Since \(aN=bN\) and \(N\subseteq K\text{,}\) we have that \(aK=bK\) by Statement 6 of Theorem 7.2.7; thus \(\phi(a)=\phi(b)\) (using Lemma 9.1.1). So \(\phibar\) is well-defined.

Next, we show that \(\phibar\) is a homomorphism. Let \(aN,bN\in G/N\text{.}\) Then

Finally, for every \(a\in G\text{,}\)

so \(\phibar \circ \Psi = \phi\text{,}\) as desired.

1. This is clear, since \(\phibar(G/N)=\phi(G)\text{.}\)

2. By Corollary 9.1.2, \(\phibar\) is one-to-one if and only if \(\Ker \phibar=\{N\}\text{.}\) But

   \begin{align*} \Ker \phibar\amp =\{aN\in G/N \,:\, \phibar(aN)=e_{G'}\}\\ \amp =\{aN\in G/N \,:\, \phi(a)=e_{G'}\}\\ \amp =\{aN\in G/N \,:\, a\in \Ker\phi\}. \end{align*}

   So \(\Ker \phibar = \{N\}\) if and only if

   \begin{equation*} \{aN\in G/N\,:\,a\in \Ker\phi\}=\{N\}, \end{equation*}

   in other words if and only if \(aN=N\) for all \(a\in \Ker\phi\text{.}\) But

   \begin{align*} {3} \amp \amp \amp aN=N \amp \amp \ \ \text{ for all \(a\in \Ker\phi\) }\\ \amp \Leftrightarrow \amp \amp a\in N \amp \amp \ \ \text{ for all \(a\in \Ker\phi\) }\\ \amp \Leftrightarrow \amp \amp \Ker \phi\subseteq N\amp \amp\\ \amp \Leftrightarrow \amp \amp \Ker\phi=N,\amp \amp \end{align*}

   since we are given that \(N\subseteq \Ker \phi\text{.}\) Thus, \(\phibar\) is one-to-one if and only if \(N=\Ker\phi\text{,}\) as desired.

The tricky part is showing that the operation is well-defined. Suppose \(\theta_1, \theta_2, t_1\text{,}\) and \(t_2\) are in \(\R\text{,}\) with \(e^{i\theta_1}=e^{it_1}\) and \(e^{i\theta_2}=e^{it_2}\text{.}\) We want to show that

i.e., that

Now, \(e^{i\theta_1}=e^{it_1}\) and \(e^{i\theta_2}=e^{it_2}\) imply that \(\theta_1 -t_1 = 2\pi m\) and \(\theta_2-t_2 = 2\pi n\) for some \(m\) and \(n\) in \(\Z\text{;}\) hence,

Thus, \(e^{i(\theta_1+\theta_2)}=e^{i(t_1+t_2)}\text{,}\) so our operation is well-defined.

Since products of elements in \(S^1\) are in \(S^1\text{,}\) we have that \(S^1\) is a binary structure; moreover, associativity of the operation in \(S^1\) follows from the associativity of addition in \(\R\text{.}\) Moreover, \(e^{i0}=1\) clearly acts as an identity element in \(S^1\text{,}\) and if \(e^{i\theta}\in S^1\text{,}\) then \(e^{i\theta}\) has inverse \(e^{i(-\theta)} \in S^1\text{.}\) So \(S^1\) is a group under the described operation.

We will define an epimorphism \(\phi\) from \(\R\) to \(S^1\) with \(\Ker \phi=\Z\text{;}\) then we'll have \(\R/\Z \simeq S^1\text{,}\) by the First Isomorphism Theorem.

Define \(\phi:\R \to S^1\) by \(\phi(r)=e^{i2\pi r}\text{.}\) We have that \(\phi\) is a homomorphism, since for every \(r,s\in \R\text{,}\) we have

Moreover, \(\phi\) is clearly onto, since if \(e^{i\theta}\in S^1\text{,}\) then

Finally, \(\Ker\phi=\Z\text{:}\) indeed,

Thus, \(\Ker \phi = \Z\text{,}\) and hence \(\R/\Z\simeq S^1\text{,}\) as desired.