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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.

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$f\colon\mathbf Z\to G$, 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 $\mathbf C^*$.

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  • $\begingroup$ 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. $\endgroup$ – Bernard Aug 17 '15 at 10:20
  • $\begingroup$ why is this constraint $n\cdot f(1\bmod n)=0$? $\endgroup$ – Foggy Aug 17 '15 at 10:45
  • $\begingroup$ Because in $\mathbf Z/n\mathbf Z$ we have $n\cdot 1=0$ by definition. $\endgroup$ – Bernard Aug 17 '15 at 10:55
  • $\begingroup$ 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 $n$. $\endgroup$ – Bernard Aug 17 '15 at 11:18
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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.

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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...

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