1. ## Find Isomorphisms

Show that:

1. The additive group $\displaystyle \mathbb {R} / \mathbb {Z}$ is isomorphic to the multiplicative group of all complex numbers of modulus 1.

2. The additive group $\displaystyle \mathbb {Q} / \mathbb {Z}$ is isomorphic to the multiplicative group of all complex roots of unity.

Okay, so I understand that I need find a mapping between these groups, prove they are well-defined, linear, and one to one and onto, but would anyone please explain to me how the complex numbers of mod 1 and roots of unity going to behave?

Complex numbers of mod 1, wouldn't that be any combination of a(i)?

Roots of unity, well, I know how it would work in integers, but complex, I'm a bit lost on that.

Show that:

1. The additive group $\displaystyle \mathbb {R} / \mathbb {Z}$ is isomorphic to the multiplicative group of all complex numbers of modulus 1.

2. The additive group $\displaystyle \mathbb {Q} / \mathbb {Z}$ is isomorphic to the multiplicative group of all complex roots of unity.
Define $\displaystyle \theta: \mathbb{R}\to \mathbb{C}^{\times}$ by $\displaystyle \theta ( a) = e^{\pi i a}$.
Now use the fundamental homomorphism theorem.

Similarly with $\displaystyle \mathbb{Q}$.

And the race is on ....! Who will post first? TPH or NonCommAlg. Egg on my face if both pull out. It's neck-and-neck at the moment.
What is this?

Show that:

1. The additive group $\displaystyle \mathbb {R} / \mathbb {Z}$ is isomorphic to the multiplicative group of all complex numbers of modulus 1.
let $\displaystyle \mathbb{C}_1$ be the multiplicative group of all complex numbers of modulus 1. define $\displaystyle f: \mathbb{R} \longrightarrow \mathbb{C}_1$ by $\displaystyle f(x)=e^{2x\pi i}.$ clearly $\displaystyle f$ is a homomorphism. now if $\displaystyle z \in \mathbb{C},$ then $\displaystyle z=\cos \theta +i \sin \theta = e^{i \theta}=f(\theta/2\pi).$

so $\displaystyle f$ is onto. also $\displaystyle x \in \ker f \Longleftrightarrow e^{2x \pi i}=1 \Longleftrightarrow 2x \pi i = 2k \pi i, \ \text{for some} \ k \in \mathbb{Z} \Longleftrightarrow x=k \in \mathbb{Z}.$ so $\displaystyle \ker f = \mathbb{Z}. \ \ \ \ \square$

2. The additive group $\displaystyle \mathbb {Q} / \mathbb {Z}$ is isomorphic to the multiplicative group of all complex roots of unity.
let $\displaystyle \mathbb{C}_2$ be the the multiplicative group of all complex roots of unity. define $\displaystyle g: \mathbb{Q} \longrightarrow \mathbb{C}_2$ by $\displaystyle g(x)=e^{2x \pi i}.$ the same as above $\displaystyle g$ is a homomorphism and $\displaystyle \ker g = \mathbb{Z}.$ now choose $\displaystyle z \in \mathbb{C}_2.$ then we have

$\displaystyle z=e^{2k \pi i/n}$ for some integers $\displaystyle k,n.$ hence $\displaystyle z=g(k/n).$ so $\displaystyle g$ is onto. Q.E.D.