Kernel of homomorphism on unit circle S1

Question

Let $f : S^1 \to S^1$ be defined such that $f(z) = z^2$, where $z$ is a complex number. It's easy to check that this is a homomorphism on $S^1$. However, how would you find the kernel and the coset of $\ker(f)$? From explicit computation, I found that $\ker(f)$ is just $1$, which is when $\theta$ is $2(\pi)(k)$.

Answer 1

I assume this is Exercise $2.2.1$ from Stillwell's Naive Lie Theory? I took this awhile back at Cornell and have since decided to write a solutions manual to this because I enjoyed it so much. Here is my solution:

First, we check that this is indeed a homomorphism. Let the mapping $z \mapsto z^2$ be denoted by $\varphi(z)$. Every element of $\mathbb{S}^1$ can be put in the form $e^{i\theta}$. Now notice that $$ \varphi(e^{i \theta_1}e^{i\theta_2})=\varphi(e^{i(\theta_1+\theta_2)})=e^{2i(\theta_1+\theta_2)}=e^{2i\theta_1}e^{2i\theta_2}=\varphi(e^{i\theta_1})\varphi(e^{i\theta_2}) $$ so that $\varphi$ is a homomorphism. So the kernel is composed of all $z \in \mathbb{S}^1$ such that $z^2=1$. There are many ways of calculating this, one way is to let $z=a+ib$, where $a^2+b^2=1$ (as $z \in\mathbb{S}^1$), then note that $$ z^2=a^2-b^2+2iab=1+0i $$ Implies that $a^2-b^2=1$ and $2ab=0$. This means either $a=0$ or $b=0$. If $a=0$ then $-b^2=1$, but since $b \in \mathbb{R}$, this is not possible. Then $b=0$ and $a^2=1$ meaning that $a=1$ and $a=-1$ are the only solutions. Another way is to observe that multiplying $z$ by itself doubles the angle $z$ makes with the $x$-axis. We need $z$ to end up at the point $(1,0)$ on the unit circle. Obviously, $z=1$ is a solution. Notice that $z=e^{i\pi}=-1$ is also a solution.

In any case, we have shown that $\text{ker }\varphi=\{\pm 1\}$. Now we look at $\mathbb{S}^1/\text{ker }\varphi$, i.e. $$ \mathbb{S}^1/\text{ker }\varphi=\{z \cdot \text{ker }\varphi \;|\; z \in \mathbb{S}^1\} $$ where $\cdot$ is the group operation in $\mathbb{S}^1$-complex multiplication. So for each $z \in \mathbb{S}^1$, when we do $z \cdot \text{ker }\varphi$, we get $\{z,-z\}$. So $$ \mathbb{S}^1/\text{ker }\varphi=\left\{\{z,-z\}\;|\; z\in \mathbb{S}^1 \right\} $$ Therefore, the cosets of the kernel consists of all the pairs of $z \in \mathbb{S}^1$ with its reflection through the origin.