I'm just starting in abstract algebra, and an exercise in a book sparked the following question.
Let's say we want to define a homomorphism $\varphi$ from a cyclic group $G$ (say, $\mathbb{Z}/d\mathbb{Z}$) to some arbitrary group $H$ (say, $S_n$ for $n \geq d$).
Since $G$ is a cyclic group, $\varphi$ can be defined by choosing an element $h \in H$, stating that $\varphi(g_G) := h$ (where $g_G$ is some generator in $G$) and requiring $\varphi({g_G}^k) = h^k$.
Now, clearly, we also need some requirements on $h$: for instance, for the aforementioned example groups, choosing $\varphi([1]) := e_{S_n}$ or $\varphi([1]) := r_d$ (where $e_{S_n}$ is the identity permutation, and $r_d$ is the "rotation" of the first $d$ elements) clearly works, while $\varphi([1]) = \{ \text{swapping first two elements} \}$ does not (at least, for odd $d$).
I could only think of the requirement that $h^d = e_H$. Is it sufficient? Is it necessary?
Best Answer
If $G$ is a (finite) cyclic group of order $d$, with generator $g_0$, and $\varphi\colon G\to H$ is a group homomorphism, then $\varphi$ is determined by $\varphi(g_0)$, which must be an element $h\in H$ satisfying $h^d=1$, because $$ 1=\varphi(1)=\varphi(g_0^d)=\varphi(g_0)^d=h^d $$
Conversely, if you take $h\in H$, with $h^d=1$, then you can define a group homomorphism $\psi\colon \mathbb{Z}\to H$ by $\psi(n)=h^n$. Since $\psi(d)=h^d=1$, we see that $d\mathbb{Z}\subseteq\ker\psi$, so by the homomorphism theorems, $\psi$ induces a unique group homomorphism $\bar{\psi}\colon\mathbb{Z}/d\mathbb{Z}\to H$ such that, for every $n\in\mathbb{Z}$, $\bar{\psi}(n+d\mathbb{Z})=\psi(n)=h^n$.
Now consider the isomorphism $\alpha\colon \mathbb{Z}/d\mathbb{Z}\to G$ sending $1+d\mathbb{Z}$ to $g_0$ and you get $$ \varphi=\bar{\psi}\circ\alpha^{-1} $$ that satisfies $\varphi(g_0)=h$.