I am reading the answer here: https://math.stackexchange.com/a/577942/130018
Now let $G$ be a nonabelian group of order $20$. By Sylow’s Theorem, $n_5 = 1$, so that $G$ has a unique (hence normal) Sylow $5$-subgroup $H \cong \mathbb{Z}_5$. Now let $K$ be any Sylow $2$-subgroup of $G$. By Lagrange, we have $H \cap K = 1$, so that $G = HK$. By the recognition theorem for semidirect products, $G \cong H \rtimes_\varphi K$ for some $\varphi : K \rightarrow \mathsf{Aut}(H)$. Evidently, classifying the nonabelian groups of order $20$ is equivalent to determining the nonisomorphic groups constructed in this manner. To that end, let $H = \mathbb{Z}_5 = \langle y \rangle$. Note that $\mathsf{Aut}(H) = \langle \alpha \rangle \cong \mathbb{Z}_4$; where $\alpha(y) = y^2$.
I don't understand the last sentence, the notation $\alpha(y)$ or why $\alpha(y) = y^2$ (how can a generator of a group be a mapping?) and why not $y^4$ (or some other arbitrary power).
Best Answer
Note that any homomorphism of $\mathbb Z_5$ is determined by the image of a generator. It is an automorphism if that image is non-trivial. To be a little more explicit about things:
If $y$ is a generator, $\alpha$ is an automorphism and $\alpha(y)= y^2$ then $\alpha (y^r)=y^{2r}$
If we apply $\alpha$ twice, $\alpha^2(y)=y^4$, three times gives $\alpha^3(y)=y^8=y^3$ and $\alpha^4(y)=y^{16}=y$ so that $\alpha^4$ is the identity automorphism.