# Why is the image of a finite group under a nontrivial homomorphism into $\mathbb{C}-\{0\}$ a set of roots of unity?

Let G be a finite group and $\phi$ a homomorphism from G to $\mathbb{C}-\{0\}$ such that $\phi(G)\neq \{1\}$. It is claimed, in a book Im studying, that $\phi(G)=\{\zeta_n^r : 1\leq r \leq n\}$, $n\geq 2$. Prove this.

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Your title says "cube roots" but you apparently meant "$n^\text{th}$ roots". It is generally preferred that in your question you ask a question rather than give directions. – Jonas Meyer Mar 21 '11 at 23:17

If $g\in G$ then $g^n=1$ for some $n$ (for instance, $|G|$). Then $1=\phi(1)=\phi(g^n)=\phi(g)^n$. This shows that the image of $\phi$ consists of roots of unity. Since there are only a finite number of them, you can take them to be powers of a single primitive root of unity.
That proves that containment holds for some $n$. Taking the least such $n$ will yield equality. (Edit: This comment was posted when this answer consisted only of the first line.) – Jonas Meyer Mar 21 '11 at 23:20
Taking $n=|G|$ proves equality right? Scratch that, I get it now. – Jason Smith Mar 21 '11 at 23:32
@Jason: No. $n=|G|$ will never work unless $G$ is cyclic, and even if $G$ is cyclic this would only hold if $\phi$ is injective. E.g., you could have $\phi:\mathbb{Z}/4\mathbb{Z}\to\{-1,1\}$. Consider $(\mathbb{Z}/2\mathbb{Z})^2$ for a case where the order of the group must be larger than $n$. – Jonas Meyer Mar 21 '11 at 23:37