# All homomorphisms from $\Bbb{Z}$ and $\Bbb{Z}/n\Bbb{Z}$ into $(\Bbb{C},+)$, $(\Bbb{C}^{\times},\cdot)$ and $(S^1,\cdot)$.

The question is to find all group homomorphisms from $\Bbb{Z}$ and $\Bbb{Z}/n\Bbb{Z}$ into $(\Bbb{C},+)$, $(\Bbb{C}^{\times},\cdot)$ and $(S^1,\cdot)$.

How should this be tackled? What should be in your arsenal in order to confront this quesion?

I have read about homomorphisms, all I know is that if $\phi:\ G\ \longrightarrow\ G'$ is a group homomorphism, then

1. $\phi(e)=e'$
2. $\ker{(\phi)}= \{x\in G\,: \phi(x)=e'\}$
3. The order of the image of an element divides order of element.

But to find homomorphisms between groups or tell their number looks tricky to me. Please explain.

$f\colon\mathbf Z\to G$, where $G$ is any abelian group:

As $f(n)=f(n\cdot 1)=nf(1)$, $f$ is defined by the sole value of $f(1)$, which is arbitrary. Hence $\;\operatorname{Hom}_{\mathbf Z}(\mathbf Z,G)\simeq G$ (the index $\mathbf Z$ in Hom stresses the fact that abelian groups are $\mathbf Z$-modules).

$f\colon\mathbf Z/n\mathbf Z\to G$, where $G$ is any abelian group:

Such a morphism comes from a morphism from $\mathbf Z$ to $G$, that vanishes on the subgroup $n\mathbf Z$. What ‘vanishes’ means depend on the notation for the group law: if $G$ is an additive group, this means $f(n)=0$, for a group noted multiplivatively, it is $f(n)=1$.

In a concrete way, a morphism $Z/n\mathbf Z\to\mathbf C$ is defined by the value of $f(1\bmod n)$ subject to the constraint $nf(1\bmod n)=0$, whence $f(1\bmod n)=0$: the only morphism is the null-morphism.

$f:Z/n\mathbf Z\to\mathbf C^\times$ is also defined by the value of $f(1\bmod n)$, but subject to the constraint $f(n\cdot 1\bmod n)=f(1\bmod n)^n=1$, i.e. $f(1\bmod n)$ is an $n$th root of unity.

This is indeed true for any homorphism $f$ with values in a commutative ring $A$:

• Consider the composition $\;\mathbf Z\xrightarrow{\pi} \mathbf Z/n\mathbf Z\xrightarrow{f} A$ and set $g=f\circ\pi$. We have $g(n)=0=f(n\cdot1\bmod n)$.
• Conversely, for any $g\colon\mathbf Z\to A$ such that $g(n)=0$, we can define $f\colon\mathbf Z/n\mathbf Z\to A$ by $f(1\bmod n)=g(1)$.

In other words: $$\operatorname{Hom}_{\mathbf Z}(Z/n\mathbf Z,\mathbf C^\times)\simeq \mathbf U_n$$ and similarly $$\operatorname{Hom}_{\mathbf Z}(Z/n\mathbf Z,S^1)\simeq \mathbf U_n,$$ since $S^1$ is a subgroup of $\mathbf C^*$.

• Actually, it is any homorphism $f\colon\mathbf Z/n\to A$ ($A$ any commutative ring) that is defined by the value of $f(1\bmod n)$ subject to $n\cdot f(1\bmod n)=0$. I've added some details. – Bernard Aug 17 '15 at 10:20
• why is this constraint $n\cdot f(1\bmod n)=0$? – Foggy Aug 17 '15 at 10:45
• Because in $\mathbf Z/n\mathbf Z$ we have $n\cdot 1=0$ by definition. – Bernard Aug 17 '15 at 10:55
• That's in the definition of $\mathbf Z/n\mathbf Z$: $\;0\equiv n\equiv 2n\equiv\cdots$ modulo $n$. The congruence class of $0$ is the set of multiples of $n$. – Bernard Aug 17 '15 at 11:18

Some hints:

• For $\phi : \mathbb{Z}/n\mathbb{Z} \to \mathbb{C}^{\times}$ or $S^1$. In both cases, you need $\phi(0)=1$, and $\phi(1)^n = \phi(n \cdot 1) = \phi(0) = 1$ which very much limits your options.
• Likewise, for $\phi : \mathbb{Z}/n\mathbb{Z} \to \mathbb{C}^+$, you need $\phi(0)=0$ and $n\phi(1)=0$. This again very much limits your options.
• For $\phi : \mathbb{Z} \to \mathbb{C}^{\times}$ or $S^1$ you have $\phi(0)=1$ and $\phi(n)=\phi(1)^n$ for all $n \in \mathbb{Z}$; thus $\phi$ is determined by the value of $\phi(1)$.
• For $\phi : \mathbb{Z} \to \mathbb{C}^+$ you have $\phi(0)=0$ and $\phi(n)=n\phi(1)$, so again $\phi$ is determined by the value of $\phi(1)$.

This information allows you to classify all the homomorphisms you're trying to classify.

The three facts you know about group homomorphisms are sufficient to answer the question, if you are familiar with the groups mentioned. I will sketch how to find all group homomorphisms from $\Bbb{Z}$ to $(\Bbb{C},+)$ as an example. For the other groups a similar approach will work.

If $\varphi:\ \Bbb{Z}\ \longrightarrow\ (\Bbb{C},+)$ is a group homomorphism, then $\varphi(0)=0$. Because $\Bbb{Z}$ is generated by $1\in\Bbb{Z}$, for all $n\in\Bbb{Z}$ we have $$\varphi(n)=n\cdot\varphi(1),$$ so if $\varphi(1)=\varphi'(1)$ for a pair of group homomorphisms from $\Bbb{Z}$ to $(\Bbb{C},+)$, then $\varphi=\varphi'$. Also, for every $c\in\Bbb{C}$ the map $$\varphi_c:\ \Bbb{Z}\ \longrightarrow\ (\Bbb{C},+):\ n\ \longmapsto\ n\cdot c,$$ is easily verified to be a group homomorphism with $\varphi_c(1)=c$. This shows that the map $$\operatorname{Hom}(\Bbb{Z},(\Bbb{C},+))\ \longrightarrow\ \Bbb{C}:\ \varphi\ \longmapsto\ \varphi(1),$$ is a bijection, and hence that all homomorphisms from $\Bbb{Z}$ to $(\Bbb{C},+)$ are of the form $\varphi_c$ for some $c\in\Bbb{C}$.

For the other groups similar arguments can be used, though there is still quite some work to be done...