How to solve Fundamental Theorem of Homomorphisms problems

Question

I found a video on youtube that shows how to use the fundamental theorem to solve these type of questions [https://www.youtube.com/watch?v=Pal391yUoWE] and it says to use four steps, the first two being to define a map G->H that is a homomorphism. Where H is the set after the congruent/isomorphic symbol. But I can't seem to create that for any of these problems. Is there a different way of doing these problems?

Answer 1

Recall that the first isomorphism theorem states:

If $\varphi: G \to H$ is an onto homomorphism, then $G/\ker(\varphi) \cong H$.

Now let's look at a), in which $G = \Bbb C$ and $H = \Bbb R$. As you've read, we should indeed be looking for a homomrphism from $\varphi: \Bbb C \to \Bbb R$, but that doesn't seem like a lot to go on. What you should also pay attention to is the fact that the kernel should match up too. In this case, we should have $\ker \varphi = U$.

Let's take that statement apart for a second. In this case, we have $$ \ker \varphi = \varphi^{-1}(e_H) = \varphi^{-1}(1) = \{x \in \Bbb C: \varphi(x) = 1\} $$ which means that for our map $\varphi$, we should have $\varphi(x) = 1 \iff x \in U$. Now, what "natural" map from $\Bbb C$ to $\Bbb R$ is such that $\varphi(x) = 1$ exactly when $x$ is in the unit circle?

Spoiler:

How about $\varphi(x) = |x|$?

Check that this map is indeed a homomorphism. Since this function is a homomorphism is whose kernel is $U$, we can now use the fundamental theorem.

Now, try all this for b). What "natural" map $\psi:\Bbb C \to U$ is such that $\psi(x) = 1$ whenever $x$ is a real number? What's the natural way to take a complex number and "make it" a complex number with length $1$?

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That second question is quite a bit trickier. To construct the homomorphism you want, we can start by noting that every rational number can be written in the form $$ q = 2^n\frac{a}{b} $$ where both $a$ and $b$ are odd. Define $\varphi(q) = n$.

Answer 2

For the first one, you need a homomorphism $\varphi : G \rightarrow \mathbb{R}^{+}$ which has precisely all the roots of unity as kernel, i.e. $\varphi(u) = 1$ if and only if $u \in U$. An obvious candidate might be to take $\varphi(x) = |x|$ which is clearly a homomorphism that satisfies these properties.

If you want to do it differently, you will still have to provide a bijective homomorphism in one direction or the other. Since providing a homomorphism $G/U \rightarrow \mathbb{R}^{+}$ is at least as hard as providing a homomorphism $G \rightarrow \mathbb{R}^{+}$ and since we assume there is a difficulty there, you could try to go in the other direction. You can define $f : \mathbb{R}^{+} \rightarrow G/U$ which for example sends and $x$ to its class modulo $U$. Checking that this is a bijective homomorphism is not that hard. I advise you to make the first reasoning though, since it is used constantly in group theory (and commutative algebra, topology...)

Answer 3

Hint:

Consider the exponential form of non-zero complex numbers: $\;z=r\mathrm e^{i\theta}$, and the projections: \begin{align} \mathbf C^*&\longrightarrow \mathbf R^+,&\mathbf C^*&\longrightarrow \mathbf U,\\ z=r\,\mathrm e^{i\theta}&\longmapsto r,&z=r\,\mathrm e^{i\theta}&\longmapsto \mathrm e^{i\theta}. \end{align}