The commutator subgroup of real quaternion division ring Let $\mathbb{H}$ be the ring of real quaternions and let $\mathbb{H}^*$ be the set of units in $\mathbb{H}$. Denoted by $[\mathbb{H}^*,\mathbb{H}^*]$ the commutator subgroup of $\mathbb{H}^*$. I want to find the properties of $[\mathbb{H}^*,\mathbb{H}^*]$. I merely know that every element $x=a+bi+cj+dk\in[\mathbb{H}^*,\mathbb{H}^*]$ has $a^2+b^2+c^2+d^2=1$. I wonder if such $x$ can be written a commutator $yzy^{-1}z^{-1}$ in which  $z$ is not conjugate to $z^{-1}$.
 A: Note that in general, the set $\{[y,z]\mid y,z\in G\}$ of commutators of a group $G$ is not a subgroup; the commutator subgroup is generated by this set. For the unit quaternion group $S^3$ it will turn out the set of commutators is the full commutator subgroup, which is all of $S^3$.
First we will claim $S^3$ is perfect, i.e. $S^3=[S^3,S^3]$. Note that since ${}^g[y,z]=[{}^gy,{}^gz]$ (where ${}^gu:=gug^{-1}$ is conjugation), we can say the set of commutators is a normal subset, and the commutator subgroup is a normal subgroup.
Proposition I. The only proper nontrivial normal subgroup of $S^3$ is $S^0=\{\pm1\}$.
One way to show this is by showing $\mathrm{SO}(3)$ (the group it double-covers) is simple, which you can do with some geometry. Thus it suffices to show the commutator subgroup has more than just $\pm1$. We can go further than this, though:
Proposition II. $S^3=[S^2,S^2]$. That is, every versor (unit quaternion) is the commutator of two unit vectors (pure imaginary quaternions).
Proof One. Let's lay some groundwork first.
Lemma A. Note that every versor has a polar form $e^{\theta\mathbf{u}}$ for a convex angle $0\le\theta\le\pi$, where $\mathbf{u}$ is unique for nonreal versors (i.e. $\ne\pm1$) and $\theta$ is always unique.
Lemma B. The effect of conjugating by $e^{\theta\mathbf{u}}$ is leaving the real part of a quaternion unchanged and rotating its imaginary part (in 3D space) around $\mathbf{u}$ by the double angle $2\theta$.
Lemma C. The conjugacy classes of $S^3$ are the spheres $\exp(\theta S^2)=\{e^{\theta\mathbf{u}}\mid\mathbf{u}\in S^2\}$ for convex angles $0\le\theta\le\pi$ (which degenerate to singular points $\{+1\}$ when $\theta=0$ and $\{-1\}$ when $\theta=\pi$ but are otherwise geometrically round $2$-spheres). The fact $S^3$ is a disjoint union of these spheres is an illustration of the fact $S^3$ is the topological join $S^0\star S^2$. This follows from the previous two lemmas.
Consider the map $S^2\times S^2\xrightarrow{[\cdot,\cdot]} S^3\xrightarrow{\mathrm{Re}}[-1,1]$. The continuous image of a connected subset is connected, so the image is a subinterval of $[-1,1]$. Pairs of parallel vectors map to $+1$ while pairs of perpendicular vectors map to $-1$, so the range is all of $[-1,1]$, which is the range of $\cos$ applied to convex angles $\theta\in[0,\pi]$. This means $[S^2,S^2]$ has a representative of each conjugacy class, but it's also a normal subset, so it must be all of $S^3$.
Proof Two. We can be more constructive. Split the commutator map into two maps:

*

*the twisty map $S^2\times S^2\to S^2\times S^2$ given by $(\mathbf{y},\mathbf{z})\mapsto (\mathbf{yzy}^{-1},\mathbf{z}^{-1})$,

*the multiplication map $S^2\times S^2\to S^3$ given by $(\mathbf{u},\mathbf{v})\mapsto\mathbf{uv}$).

Lemma D. The twisty map is onto. If $(\mathbf{u},\mathbf{v})\in S^2\times S^2$ is arbitrary, we can set $\mathbf{z}=\mathbf{v}^{-1}=-\mathbf{v}$ and then pick $\mathbf{y}$ as a midpoint of a geodesic arc from $\mathbf{z}=-\mathbf{v}$ to $\mathbf{u}$. If $\mathbf{u}\ne\mathbf{v}$ the only choice is $\mathbf{y}=(-\mathbf{v}+\mathbf{u})/\|-\mathbf{v}+\mathbf{u}\|$ but otherwise if $\mathbf{u}=\mathbf{v}$ we can pick any $\mathbf{y}$ on the great circle perpendicular to it.
Lemma E. The multiplication map is onto. The product of unit vectors is $\mathbf{uv}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$. For this to equal $x=e^{\theta\mathbf{w}}=\cos\theta+\sin\theta\,\mathbf{w}$, we simply need to pick $\mathbf{u}$ and $\mathbf{v}$ a supplementary angle $\pi-\theta$ apart from each other in the great circle orthogonal to $\mathbf{w}$, oriented appropriately.
The lemmas imply $S^3=[S^2,S^2]$.
