Understanding proof about a group and homomorphism

Question

I have a book about group theory and there was the following question:

Let $G$ be a set of all the real matrices in the following form: $\begin{pmatrix}a & b\\ -b & a \end{pmatrix}$ when $a^2+b^2>0$.

1. Prove that $G$ is a group.
2. Prove that $G\cong (C^\times,\cdot )$.

I successfully proved that $G$ is a group. Now I'm trying to prove the second sub-question. In the book they suggested to declare the following function:

$$ f:\begin{pmatrix}a & b\\ -b & a \end{pmatrix} \to a+ib$$

Also they wrote "obviously $f$ is bijection", and then they proved the Homomorphism equation. The only part that I didn't understand is why $f$ is a bijection, and why it is so obvious? How can I prove it formally?

Answer 1

Since $\mathbb{C}^\times=\mathbb{C}\setminus\{0\}$ and since each $z\in\mathbb{C}\setminus\{0\}$ can be written in one and only one way as $a+bi$ with $a$ and $b$ not both equal to $0$ ($\iff a^2+b^2>0$), it is clear that $f$ is a bijection.

Answer 2

To see that the map is a bijection note that the kernel consists precisely of the identity matrix and hence the map is injective. Indeed if $f(A)=1=a+ib$ for some matrix $A=\begin{bmatrix} a& b\\ -b& a \end{bmatrix} $, then we have that $a=1$ and $b=0$. Conversely given any $z\in \mathbb{C}\setminus {0}$, let $a=\text{Re}(z)$, and $b=\text{Im}(z)$, so that $$ \begin{bmatrix} a& b\\ -b& a \end{bmatrix}\to z $$ whence the map is surjective.

Answer 3

For every complex number $\mathbb{C}-\{0\} \ni x:=a+bi$,(hence $a \neq 0$ or $b \neq 0$). Consider the matrix $$y=\begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix}$$, The determinant of this matrix of $a^2+b^2 >0$ and it gets map to $a+bi$ by $f$(that is it! $f(y)=x$). So this shows surjectivity!

For injectivity, we have if $0\neq a+bi=f(A)=f(B)=c+di \neq 0$, then $a=c, b=d$. so $$A=\begin{bmatrix} a & b \\ -b & a \\ \end{bmatrix}=\begin{bmatrix} c & d \\ -d & c \\ \end{bmatrix}=B$$ And so $f$ is bijective!