(a) Using exercises 4 and 7(c) of §22, $A\cdot b$ is closed and there are two disjoint open sets $U_b$ and $V_b$ containing given $c\notin A\cdot B$ and $A\cdot b$ , respectively. We show there exists some neighborhood $W_b$ of $b$ such that $A\cdot W_b\subseteq V_b$ . Indeed, for each $a\in A$ there is a neighborhood $K_a$ of $b$ such that $a\cdot b\in a\cdot K_a\subseteq V_b$ (using the fact that $V_b$ is open and the multiplication is a homeomorphism, exercise 4 of §22), and their $W_b$ union is the neighborhood of $b$ such that $A\cdot b\subseteq A\cdot W_b\subseteq V_b$ . Now, $\{W_b\}$ cover $B$ and there is a finite subcovering. The corresponding finite intersection $U_b$ is the neighborhood of $c$ disjoint from $A\cdot B$ .(b) If $A$ is closed in $G$ and $xH\notin p(A)$ then $x\notin A\cdot H$ which is closed by (a). And there is a neighborhood $U$ of $x$ such that it does not intersect $A\cdot H$ . Then, $p(U)$ is open (exercise 5(c) of §22), contains $xH$ and does not intersect $p(A)$ . Therefore, $p(A)$ is closed.(c) $p$ is continuous and closed (by (b)), moreover, $p^{-1}(xH)=xH$ is compact as the multiplication is a homeomorphism. Therefore, by exercise 12, $G$ as the preimage of $G|H$ is compact.