Remark

In the notation of Proposition 7, the fact that the inverse image of a subgroup of $H$ is a subgroup of $G$ follows from the following more general fact.

If $A$ and $B$ are subsets of $H$ and $f$ is surjective, then

Obviously $f^{-1}(A)\cdot f^{-1}(B)\subset f^{-1}(A\cdot B)$; on the other hand, if $z\in f^{-1}(A\cdot B)$, there exists $a\in A$ and $b\in B$ such that $f(z)=ab$; as $f$ is surjective, there exists $x\in G$ such that $f(x)=a$; writing $y=x^{-1}z$, $f(y)=a^{-1}f(z)=b$ and $z=xy$, whence $z\in f^{-1}(A)\cdot f^{-1}(B)$. The relation $x\in f^{-1}(A^{-1})$ is equivalent to $f(x)\in A^{-1}$, hence to $f(x^{-1})\in A$, that is to $x^{-1}\in f^{-1}(A)$ and finally to $x\in f^{-1}(A)^{-1}$.