How Does G/N Relate to Complex Numbers of Absolute Value 1?

[fk378]

Homework Statement

Let G be the group of real numbers under addition and let N be the subgroup of G consisting of all the integers. Prove that G/N is isomorphic to the group of all complex numbers of absolute value 1 under multiplication.

Hint: consider the mapping f: R-->C given by f(x)=e^[2pi(ix)]

The Attempt at a Solution

So this says that a subgroup of Z is normal in R.
G/N is the quotient group of left cosets of N in G.
And I want to prove that G/N is isomorphic to (a+bi)(c+di) <---not sure if this is what I want to prove...but if it is then...it equals ac+adi+bci-bd= +/- 1
Which implies
ac+adi+bci-bd=ac-bd=+/- 1

Am I thinking of this right so far?

I'm not sure how to use the hint.

 

[CompuChip]

The hint gives you a function from R to C. You want a function from R/N to some subset of C, however. Can you construct such a function from the given one? Can you prove it is an isomorphism?

 

[fk378]

The given function is cos2(pi)x+isin2(pi))x...I'm not sure how to construct a mapping for this...I'm pretty lost.

 

[morphism]

Do you notice anything special about f, such as say, if it's a homomorphism (between appropriate groups)..?

 

[fk378]

Between appropriate groups? But how is G/N defined in this problem...I don't understand.
I know e^(2pi)(ix) = cos2(pi)x+isin2(pi)x...

 

[morphism]

f is a homomorphism between G and the group of all complex numbers of absolute value 1 under multiplication.

What's its kernel?

 

[fk378]

The kernel is all the elements of the image of the mapping?

 

[morphism]

No. The kernel is the set of elements in G that get mapped to the identity.

 

[Mopho]

What it means for a complex number to have absolute value 1, is: given $a+bi \in \C$, we say that $a+bi$ has absolute value $1$ if $a^2+b^2 =1$.

We can see that the function in your hint, $e^{2\pi (ix)} = \cos{2 \pi x} + i\sin{2 \pi x}$. This function is well-defined, as it maps elements of $G/N$ to elements of $\C^*$ ($\C^*$ is the group of complex numbers of absolute value 1) with an absolute value of $1$ because $\cos^2{x}+\sin^2{x} =1$ for all $x$

Now:

To show $\phi$ is a homomorphism, let $x,y \in G/N$, and
\begin{align*}
\phi(x+y) &= e^{2 \pi [i(x+y)]}\\
&= e^{2 \pi(ix) + 2 \pi(iy)}\\
&=e^{2 \pi(ix)}e^{2 \pi(iy)}\\
&=\phi(x)\phi(y)
\end{align*} and so, we see $\phi$ is a homomorphism.

To show something that is injective or surjective is fairly straightforward so I'll leave that to you.