Unit Hurwitz quaternions forms a group Define Hurwitz quaternions by $\mathbb{H}=\left\{a+bi+cj+dk\in \mathbb {H} \mid a,b,c,d\in \mathbb {Z} \;{\mbox{ or }}\,a,b,c,d\in \mathbb {Z} +{\tfrac {1}{2}}\right\}.$
My question is, does unit quaternions (i.e. those quaternions $q\in\mathbb{H}$, for which $\left|q\right|=\sqrt{a^2+b^2+c^2+d^2}=1$) forms a group?
There are $24$ unit quaternions: $\pm1,\pm i, \pm j, \pm k, \frac{1}{2}\left(\pm 1\pm i\pm j\pm k\right).$
How to show for example associativity? I mean, I could check all combinations of elements or point to fact that Hurwitz quaternions satisfy associativity too, but I don't think it is the best way to go.
 A: The Hurwitz quaternions, which is also known as the binary tetrahedral group $2T$ (and I will explain this at the end), inherit associativity from quaternion multiplication itself. If $(X,\cdot)$ is any set with an associative operation $\cdot$, and $Y\subseteq X$ is a subset closed under that operation, then $(Y,\cdot)$ itself is an algebraic structure with an associative operation! But it still means you have to show $2T$ is closed under multiplication.
Of course, this also means you should have verified quaternion multiplication on $\mathbb{H}$ was associative to begin with. You can do this by first showing it is associative for certain expressions involving basis elements $\mathbf{i},\mathbf{j},\mathbf{k}$ (by symmetry you don't have to check all possible expressions, only some of them), then argue it works for all quaternions since $\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$ spans them. Or, if you instead define quaternion multiplication using dot products and cross products (this is a more "coordinate-free" and "intrinsic" definition), you would check associativity using dot product and cross product identities and geometry. (Fun!)
Now you want to show $2T$ is closed under multiplication. For this, I recommend splitting this into two facts: the Hurwitz integers are closed under multiplication, and the unit quaternions aka versors (which form a three-sphere $S^3$, just as unit complex numbers aka phasors form a unit circle $S^1$) are closed under multiplication. We can show $S^3$ is a group by using the fact that the quaternion norm is multiplicative. We can show the Hurwitz quaternions are closed under multiplication by checking the numerators of the components mod $2$ are $1+1+1+1$. (Or, arguably you could show Hurwitz quaternions are closed under multiplication by showing $2T$ is a group and a $\mathbb{Z}$-spanning set.)
Indeed, we can determine $2T$'s structure. If you denote the $3$rd root of unity
$$ \omega=\tfrac{1}{2}\big(-1+\mathbf{i}+\mathbf{j}+\mathbf{k}\big)=\exp\big(\tfrac{2\pi}{3}(\mathbf{i}+\mathbf{j}+\mathbf{k})\big), $$
you can verify $2T=Q_8\sqcup Q_8\omega\sqcup Q_8\omega^2$, where $Q_8=\{\pm1,\pm\mathbf{i},\pm\mathbf{j},\pm\mathbf{k}\}$ is the so-called "quaternion group" (I used to refer to $2T$ by $Q_{24}$, fun fact). Indeed, by checking $\omega Q_8\omega^{-1}= Q_8$ (which requires merely checking for $\mathbf{i},\mathbf{j},\mathbf{k}$ since $\pm1$ are evidently central) we see $Q_8$ is normal, so indeed $2T=Q_8\rtimes C_3$ is a semidirect product, where $C_3=\langle\omega\rangle$ is cyclic of order $3$.
It's also fun to note that while the Lipschitz integers $\mathbb{Z}[\mathbf{i},\mathbf{j},\mathbf{k}]$ give you a tiling of tesseracts, the diagonals of the hypercubes have length $\sqrt{1+1+1+1}=2$, so if we situate spheres of diameter $1$ at integer coordinates there is exactly enough room to fit more spheres at half-integer coordinates. In other words, this is how a 4D grocer would stack 4D oranges.
The unit quaternions $S^3$ can be used to model 3D rotations. Quaternions have polar forms, and if we have a versor $p=\exp(\theta\mathbf{v})$, then $p\mathbf{x}p^{-1}$ as a function of 3D vectors $\mathbf{x}$ is a rotation around the axis $\mathbf{v}$ by an angle $2\theta$. Indeed, there is a $2$-to-$1$ onto group homomorphism $S^3\to\mathrm{SO}(3)$ (the latter being the group of $3\times3$ rotation matrices) with kernel $S^0=\{\pm1\}$ (the unit reals). The regular tetrahedron has a orientation-preserving symmetry group of order $12$ called $T\cong A_4$ achieving all even permutations of its four vertices. (The full symmetry group includes reflections and is isomorphic to $S_4$, achieving all permutations of the four vertices.) The preimage of $T$ under the spin homomorphism $S^3\to\mathrm{SO}(3)$ is the so-called binary tetrahedral group $2T$ of order $24$. We can pick a cube, centered at $\mathbf{0}$ of side-length $2$ with vertices $\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k}$, and there is an inscribed tetrahedron with alternative vertices of evenly-many minus signs. The corresponding unit quaternions in $2T$ are precisely the unit Hurwitz quaternions!
