Exercise 14.3.7

Proof. (a) $\text{∙}$ Since $e\in H,e\in K$, $e=\mathit{ee}\in \mathit{HK}$, thus $\mathit{HK}\ne \varnothing$. $\text{∙}$ Let $x=\mathit{hk},y=h^{\prime }{k}^{\prime }$ be any elements of $\mathit{HK}$, where $h,h^{\prime }\in H,k,k^{\prime }\in K$. Then

where $h^{″}=h{h}^{\prime }\in H,k^{″}=({h{}^{\prime }}^{-1}k{h}^{\prime })k^{\prime }$, and ${h{}^{\prime }}^{-1}k{h}^{\prime }\in K$, since $K$ is normal in $G$, so that $k^{″}=({h{}^{\prime }}^{-1}k{h}^{\prime })k^{\prime }\in K$. Thus $\mathit{xy}\in \mathit{HK}$ $\text{∙}$ If $x=\mathit{hk}\in \mathit{HK}$, where $h\in H,k\in K$, then

where $h^{-1}\in H$, and $h{k}^{-1}{h}^{-1}\in K$, since $K$ is normal in $G$. Thus $x^{-1}\in \mathit{HK}$.

We have proved that $\mathit{HK}$ is a subgroup of $G$. Moreover, if $x=\mathit{hk}\in \mathit{HK}$, and $g\in G$ then

where $\mathit{gh}{g}^{-1}\in H,\mathit{gk}{g}^{-1}$ in $K$, since $H,K$ are normal subgroups. Therefore $\mathit{HK}$ is a normal subgroup of $G$. (b) Assume that $H\cap K=\{e\}$. If $h\in H,$ and $k\in K$, consider $z=\mathit{hk}{h}^{-1}{k}^{-1}\in G$. Then $z=h(k{h}^{-1}{k}^{-1})$, where $h\in H,k{h}^{-1}{k}^{-1}\in H$, so that $z\in H$. Similarly $z=(\mathit{hk}{h}^{-1})k^{-1}$, where $\mathit{hk}{h}^{-1}\in K$ and $k^{-1}\in K$, so that $z\in H\cap K=\{e\}$. Thus $\mathit{hk}{h}^{-1}{k}^{-1}=e$, which proves $\mathit{hk}=\mathit{kh}$.

(c) As in part (b), assume that $H\cap K=\{e\}$. Define

$\text{∙}$ If $u=(h,k)\in H\times K,v=(h^{\prime },k^{\prime })\in H\times K$, then part (b) shows that $h^{\prime }k=k{h}^{\prime }$, thus

thus $\phi$ is a group homomorphism. $\text{∙}$ If $x$ is any element of $\mathit{HK}$, then by definition there are some $h,k,h\in H,k\in K$ such that $x=\mathit{hk}=\phi (h,k)$, where $(h,k)\in H\times K$. Therefore $\phi$ is surjective. $\text{∙}$ if $u=(h,k)\in \ker <!--\; FUNCTION\; APPLICATION\; -->\phi$, then $\mathit{hk}=e$, thus $h=k^{-1}\in H\cap K=\{e\}$, and $(h,k)=(e,e)$. This proves $\ker <!--\; FUNCTION\; APPLICATION\; -->\phi =\{(e,e)\}$, and $\phi$ injective.

We have proved that $\phi$ is a group isomorphism. □