A Question about a non trivial morphism

Question

Define $\phi : A \to \mathbb{C}^*$ a non trivial morphism. Note that a finite subgroup of $\mathbb{C}^*$ will be cyclic.

Suppose that $A$ is cyclic. Then there exists $ x \in A$ such that $A = \langle x\rangle $. Let $m$ be the cardinal of $A$ (so $m$ is also the order of $x$).

If why suppose that $\phi $ is injective then why do we have that $\phi(A) = \mathbb{U}_m$ where $\mathbb{U}_m$ denotes the group of units?

Since $A = \langle x \rangle $, we have that $\phi(A) = \langle \phi(x)\rangle $. If $m$ is the order of $x$, then $\phi(x^m) = 1 $, so $(\phi(x))^m = 1$.

Where do we use that $\phi$ is injective?

Answer 1

I suppose that by the group of units $\mathbb{U}_m$ you mean the group of $m$-th roots of unity. Consider the following example:

$A\simeq C_4$, where $C_4$ is the cyclic group of order $4$. Then $A=\langle x\rangle$ where $x^4=1$. Let $\phi:A\rightarrow\mathbb{C}^*$ be given by $\phi(x)=-1$, then $\phi$ is a homomorphism but $\phi(A)=\mathbb{U}_2\neq \mathbb{U}_4$. What happened is that $\phi$ is not injective, in fact $\phi(x)=\phi(x^3)=-1$. You are using injectivity to show that $\phi(x)$ will also have order $m$ (and not some divisor of $m$).

Answer 2

Since $e^2=e$, but the only complex numbers that are idempotent are $0$ and $1$, we know $e$ must map to $1$

Every $\phi(a)$ for $a\in A$ is a root of $x^m-1$, since $a^m=e$ in $A$, and $e$ must map to $1$.

Because $\phi$ is injective, the image of $\phi$ contains $m$ distinct elements, and so it consists precisely of all complex roots of $x^m-1$. These roots are precisely the $m$th roots of unity, $\mathbb{U}_m$.