let $\phi :(U_n, *) \rightarrow (Z_n, +_n)$ be defined by $\phi(e^{m \frac{2\pi}{n}i}) = m$ , where $m \in {0,1,\ldots,n-1}$
This is what I did to show homomorphism.
Let $\psi(m) = e^{m \frac{2\pi}{n}i}$, for $\psi (Z_n, +_n) \rightarrow (U_n, *)$
Then $\psi(x+_ny) = \psi(x) * \psi (y)$ as $e^{(x+y) \frac{2\pi}{n}i} = e^{(x) \frac{2\pi}{n}i} * e^{(y) \frac{2\pi}{n}i}$
Is this the right way to do this? Also how do I show that $\phi$ is bijective?
thanks
Well,to answer your first question, you're confusing me here. First of all,what do you mean by the group $(U_n, *)$? I assume you mean the multiplicative group of complex roots of unity in the Argand plane. Second, is $\psi(m)$ supposed to be the inverse map of $\phi :(U_n, *) \rightarrow (Z_n, +_n)$ ? If so,this doesn't work because you've only proven that the inverse of $\phi$ is a homomorphism in the opposite direction of the original map.
Before you do any of this,however,you have to prove this map is a bijection. A map is a bijection iff it is both one to one and onto. We first need to prove it is an injective (one to one) map. Consider 2 roots of unity: $z^{j}= e^{m \frac{2\pi}{j}i}$ and $z^{k}=e^{m \frac{2\pi}{k}i}$ where j,k are positive integers.Then clearly $\phi(e^{c \frac{2\pi}{j}i}) = \phi(e^{d \frac{2\pi}{k}i)}$ where c =0,1,....j-1 and d= 0,1,...k-1.Clearly,the equality is true iff j=k Now we need to show the map is onto. Since m = x (mod n) for every m in the range of $\psi(m)$ and n is an arbitrary integer root of unity, then the map is onto the positive integers mod n. So the map is bijective.
Now that we've shown the map is bijective, let $\psi=(\phi)^{-1}$. Then your proof is valid since the inverse of any isomorphism is also an isomorphism and therefore a homomorphism. Therefore,by proving $\psi=(\phi)^{-1}$ is a homomorphism, you've also proven that $\phi$ is!