# Homomorphisms from a cyclic group

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?

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 $$\varphi(g_0)=h$$.
There's an approach to this using "abstract nonsense". Your question generalizes a common pattern in abstract algebra for defining maps: do it from a free object and quotient. Indeed, any cyclic group looks like $$\mathbb Z / n \mathbb Z$$ for $$n \geq 0$$. $$\mathbb Z$$ is the free group on one generator, so a group homomorphism $$\mathbb Z \longrightarrow G$$ is just a choice of element in $$G$$ to map 1 to. By universal property of the quotient, a map $$\mathbb Z/n \mathbb Z \longrightarrow G$$ is just a map $$\phi: \mathbb Z \longrightarrow G$$ which vanishes on $$n \mathbb Z$$, i.e. an element $$g \in G$$ such that $$\phi(1) = g$$ and $$\phi(n) = g^n = e$$, so your criterion is sufficient.