Let \(G=\{(a,b)\mid a,b\in\mathbb R,~a\neq0\}\) with \((a,b)(c,d)=(ac,ad+b)\) be a group, \(K=\{(1,b)\mid b\in\mathbb R\}\). Show that \(G/K\cong\mathbb R^*\).
Proof: Let

be a map. For any \((a,b),~(c,d)\in G\), we have

thus \(\varphi\) is a homomorphism. For any \(y\in\mathbb R^*\), there exist \((\sqrt y,c)\in G\), s.t. \(\varphi((\sqrt{y},c))=y\). Thus, \(\text{Im}\varphi=\mathbb R^*\). By the First Isomorphism Theorem, we have \(G/\ker\varphi\cong\text{Im}\varphi=\mathbb R^*\). #

Let \(m\in\mathbb Z\) and \(m>1\), \(\begin{aligned}\varphi:\quad\mathbb Z&\to\mathbb Z_m\\a&\mapsto\bar a\end{aligned}\). Prove that \(\mathbb Z/\lang m\rang\cong\mathbb Z_m\).
Proof: For any \(a,b\in\mathbb Z\), we have

so \(\varphi\) is a homomorphism. Let \(x\in\ker\varphi\), i.e., \(\varphi(x)=\bar 0\), then \(x=km\in\lang m\rang,~k\in\mathbb Z\Rightarrow \ker\varphi\subseteq\lang m\rang\). For any \(x\in\lang m\rang\), we have \(x=km,~k\in\mathbb Z\), so \(\varphi(x)=\overline{km}=\bar 0\Rightarrow \lang m\rang\subseteq \ker\varphi\). Thus, \(\ker\varphi=\lang m\rang\). And \(\text{Im}\varphi=\mathbb Z_m\). By the First Isomorphism Theorem, we have \(\mathbb Z/\lang m\rang\cong\mathbb Z_m\). #

Let \(H,K\triangleleft G\), show that \(G/HK\cong(G/H)/(HK/H)\).
Proof: For any \(hk\in HK\), where \(h\in H,~k\in K\). For any \(g\in G\), we have \(g(hk)g^{-1}=(ghg^{-1})(gkg^{-1})\). Since \(H\triangleleft G\) and \(K\triangleleft G\), we have \(ghg^{-1}\in H\) and \(gkg^{-1}\in K\), thus \(g(hk)g^{-1}\in HK\). Therefore, \(HK\triangleleft G\). Since \(H\triangleleft G\) , \(HK\triangleleft G\) and \(H\subseteq HK\), by the Third Isomorphism Theorem of groups, we have