In my lecture notes I have this exercise:
Let $G$ and $H$ be cyclic groups. Having defined the operation:
$$\varphi \psi: G \rightarrow H: x \mapsto \varphi(x) \psi(x)$$ for which $H^G$ is a group and $\operatorname{Hom}(G,H)$ is a subgroup of $H^G.$ Characterize the isomorphisms in $\operatorname{Hom}(G,H)$ when they exist.
the solution goes like this: It is easily found that since $G=\langle g\rangle$,
$$\operatorname{Im}(\varphi)=\{\varphi(g^n)=(\varphi(g))^n\ |n \in \mathbb{Z}\}$$ and
$$\operatorname{Ker}(\varphi)=\{g^n | =(\varphi(g))^n = 1 \}$$
Then they argue some things I don't understand well:
if $m=\operatorname{ord}(\varphi(g))$, it follows that $\operatorname{Ker}(\varphi)=\langle g^m\rangle <G$ ----> how did they come up with it?
If $\varphi :G \rightarrow H$ is an isomorphism, since it is surjective: $\varphi(g)$ must generate $H$ --> why is that? I know H must be generated by some element of it but why does it have to be the image of the generator of G?
And since $\varphi$ must be injective then $g^m=1$, then the $\operatorname{ord}(\varphi(g)))=\operatorname{ord}(g)$.--> I don't know how they made injectivity imply that. I would argue that they both have to have the same number of elements for finite order but can't relate it to injectivity
Can someone clarify this steps?
If $x$ is an element of $\operatorname{Ker}\phi$, then $\phi(x)=1$. Since $x$ is an element of $G$ and $G$ is cyclic generated by $g$, we may write $x=g^k$ for some integer $k$. Then $\phi(g^k)=\phi(g)^k=1$. Since the order of $\phi(g)$ divides every integer $k$ such that $\phi(g)^k=1$, you have that $m$ divides $k$, hence $x=g^k\in \langle g^m\rangle$. Conversely, if $x\in\langle g^m\rangle$, then $m$ divides $k$, hence $k$ may be expressed as a product of the form $k=mu$ for some integer $u$ and then $\phi(g^k)=(\phi(g)^m)^u=1^u=1$, so that $x\in\operatorname{Ker} \phi$.
If $\phi$ is surjective, then for every element $h\in H$ there exists an element $g^k\in \langle g\rangle$ such that $h=\phi(g^k)=\phi(g)^k$, showing that every element of $H$ is in the subgroup generated by $\phi(g)$. Conversely, if some element $h$ in $H$ is in the subgroup of $H$ generated by $\phi(g)$ then there exists an integer $k$ such that $h=\phi(g)^k=\phi(g^k)$, thus $h$ is in the image of $G$ under $\phi$, and since $h$ was arbitrary chosen, $\phi$ is finally proven to be surjective.
If $\phi$ is injective, then $\operatorname{Ker}\phi=\{ 1\}$, hence $\phi(g^m)=1$ implies that $g^m=1$. Hence, the order of $g$ divides the order of its image $\phi(g)$. Since the order of the image surely divides the order of $g$, you finally get that the two orders are the same when $\phi$ is an isomorphism.