Proposition: $\mathbb{S^3}$ can be decomposed into disjoint congruent circles
Solution:The quaternions of unit length $q=(a,b,c,d)$ of unit length satisfy the equation $$ a^2 + b^2 + c^2 + d^2 =1$$
The unit quaternions also form a group under quaternion multiplication. One subgroup of this group is $\cos \theta + i \sin \theta$ which is a circle in the plane spanned by $1$ and $i$. It follows that any coset $qH$ is also a unit circle, because multiplication by quaternion of unit length is an isometry. Since the cosets $qH$ fill the whole group and are disjoint, we have a decomposition of the $3-$ sphere into unit circles
John Stillwell's book: Naive Lie theory
The point that I don't get in the above proof is how it was taken that $qH$ is a unit circle. I understand multiplication by quaternions preserves distance between points, but how it is that just preserving length and origin is enough to know that circles will get sent to circles?
Indeed, every isometry is a congruence map (and vice versa).
Clearly, the composition of isometrics is an isometry, and the composition of congruence maps is a congruence map. Also, the reflection at a plane in 3space is clearly both an isometry and a congruence map.
Let $f$ be any isometry of 3space. Pick a point $A$ and let $r_1$ be the reflection at the plane that perpendicularly bisects the line segment $Af(A)$ (or let $r_1$ be the identity map if $f(A)=A$). Note that said plane is also precisely the locus of all points having same distance from $A$ and $f(A)$. Then $r_1\circ A$ is an isometry with fix point $A$. Next pick $B\ne A$ and define $r_2$ analogously such that $r_2\circ r_1\circ f$ fixes $B$. As $A$ has same distance from $B$ and $r_1(f(B))$, it is on the picked reflection plane and hence $A$ is also a fix point of $r_2\circ r_1\circ f$. The points on line $AB$ can be characterised by distances alone, namely as points $X$ with $d(X,A)+d(X,B)=d(A,B)$ or $|d(X,A)-d(X,B)|=d(A,B)$. Therefore, all points on this line are fix points. Pick $C$ not on the line and repeat the procedure. We obtain an isometry $r_3\circ r_2\circ r_1\circ f$ that fixes $A,B,C$ (and I fact the plane defined by these points). Repeat a last time with a fourth point $D$ not in the plane $ABC$ to obtain an isometry $r_4\circ r_3 \circ r_2 \circ r_1\circ f$ that fixes $A,B,C,D$. Any other point is uniquely determined by its distances to these four points, hence must Ali be fixed by this isometry. We conclude that our composed isometry is the identity, or equivalently, $f=r_1\circ r_2\circ r_3\circ r_4$, which is a congruence map.