Let $\mathbb{H}$ be the real quaternion division ring, that is, $\mathbb{H}$ consists of all elements of the form: $a+bi+cj+dk$ in which $a,b,c,d\in\mathbb{R}$ and $i^2=j^2=k^2=-1,ij=-ji=k$ with usual addition and multiplication. Then, the set $\mathbb{H}^*$ of all nonzero elements of $\mathbb{H}$ is a group under multiplication. I am interested in the structure of $\mathbb{H}^*$. Recently, I have read this article which is interesting. In Lemma 1, the authors show that $\mathbb{H}^*$ coincides with the group generated by $\mathbb{C}^*$ and $j+1$ in which $\mathbb{C}^*$ is the set of all nonzero complex numbers as well as the group under multiplication. However, in the proof of Lemma 1, the final claim makes me unknown. Now, let us go slight the steps of the proof.
Step 1. Let $G=\langle\mathbb{C}^*,j+1\rangle$ be the group generated by $\mathbb{C}^*$ and $j+1$. Then, $G\subseteq\mathbb{H}^*$.
Step 2. If $w,z\in\mathbb{C}$ such that $|w|=|z|$ and $z+j\in G$, then $w+j\in G$. Here $|\cdot|$ is denoted by the modulus of the complex number.
Step 3. If $a\in\mathbb{C}$, then $a+j\in G$.
Step 4. Each elements of $\mathbb{H}$ is the form: $$a+bi+cj+dk=a+bi+cj+dij=(a+bi)+(c+di)j=(c+di)\left(\dfrac{a+bi}{c+di}+j\right)\in G$$ if $c+di\neq0$. Obviously, if $c+di=0$, then $a+bi+cj+dk=a+bi\in G$.
In Step 3, the authors tell us for $r>0$, let $$u=\dfrac{1-r}{1+r}+\left(\sqrt{1-\left(\dfrac{1-r}{1+r}\right)^2}\right)i.$$ Then, $|u|=1=|1|$, and since $1+j\in G$, we obtain that $u+j\in G$ by Step 2. Thus, $(1+j)(u+j)\in G$, which implies that $$\dfrac{u-1}{\overline{u}+1}+j=\dfrac{1}{\overline{u}+1}\left((u-1)+(\overline{u}+1)j\right)=\dfrac{1}{\overline{u}+1}(1+j)(u+j)\in G.$$ On the other hand, we have $$\left|\dfrac{u-1}{\overline{u}+1}\right|=\sqrt{r}.$$ Since $r>0$ was arbitrary, by our claim for all $a\in\mathbb{C}$, we have $a+j\in G$.
The bold line mentioned above makes me unknown. Please help me explain this one. Any counterexample or reference or technique is very much appreciated. Thank you in advance.
Best Answer
As we have $\frac{u-1}{\overline{u}+1}\in G$ and $\left|\frac{u-1}{\overline{u}+1}\right|=\sqrt{r}$, Step $2$ yields that every $a\in\mathbb{C}$ with $|a|=\sqrt{r}$ lies in $G$. But $r$ was arbitrary, it can be any real positive number, and the square root is a bijective function on $\mathbb{R}_{>0}$. Thus, as every complex number except $0$ has absolute value in $\mathbb{R}_{>0}$, we see that every complex number $a\in\mathbb{C}$ is of the form $|a|=\sqrt{r}$ for some $r>0$, and since $r$ was chosen arbitrarily, i.e. any $r$ works, we see $a+j\in G$.