Exercise 8.1.3

Proof. Let $\pi :G\to G∕H$ the canonical projection, and $K$ a subgroup of $G∕H$.

(a)

${\pi }^{-1}(K)\subset G$ is the pre-image of a subgroup by the group homomorphism $\pi$, so is a subgroup of $G$. Moreover $\{\bar{e}\}=\{\mathit{eH}\}\subset K$, thus $H={\pi }^{-1}(\{\bar{e}\})\subset {\pi }^{-1}(K)$. So ${\pi }^{-1}(K)$ is a subgroup of $G$ which contains $H$.

(b)

For all $x\in G$, $x\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\pi )⟺\mathit{xH}=H⟺x\in H$.

Thus $\ker <!--\; FUNCTION\; APPLICATION\; -->(\pi )=H$.

$\mathit{eH}=H$ being the identity of $G∕H$, by definition of the kernel, $H=\ker <!--\; FUNCTION\; APPLICATION\; -->(\pi )={\pi }^{-1}(\{\mathit{eH}\})$.

(c)

As $\pi$ is a mapping of $G$ in $G∕H$, ${\pi }^{-1}(G∕H)$ is the whole group $G$.