I am trying to read the Classification of p-Groups with cyclic subgroup of index p done in Cohomology of Groups by Brown. He starts his proof by construction the exact sequence, $0→\mathbb{Z}_q→G→H→0$ ,he then says that the action of $H$ on $\mathbb{Z}_q$ is given by an embedding $H$ in $\mathbb{Z}_q$*, where $\mathbb{Z}_q^*$ is the group of units of $\mathbb{Z}_q$, and i dont quite see why this statement is true.
New edit: I didnt wanna make a new question for this since its in the same proof , but can anyone enlighted me on why the image of the embedding of $H$ in the group of units is $\{1+b:b \in p^{n-1}\mathbb{Z}/p^n\mathbb{Z}\}$?
The subgroup $\Bbb{Z}/q\Bbb{Z}$ is normal in $G$, so the action of $H$ induces an automorphism of $\Bbb{Z}/q\Bbb{Z}$. Every automorphism of $\Bbb{Z}/q\Bbb{Z}$ is given by multiplication by some invertible element in $\Bbb{Z}/q\Bbb{Z}$. This yields an embedding of $H$ in $(\Bbb{Z}/q\Bbb{Z})^{\times}$ if and only if the action of $H$ on $\Bbb{Z}/q\Bbb{Z}$ is nontrivial, because $H\cong\Bbb{Z}/p\Bbb{Z}$.
Of course, if the action of $H$ on $\Bbb{Z}/q\Bbb{Z}$ is trivial then $\Bbb{Z}/q\Bbb{Z}$ is a central subgroup giving a cyclic quotient, and so $G$ is abelian. (This answers your question from a few minutes ago, now deleted).