Finally, what is $ij$? Well, for geometric reasons it must be $\lambda_{ij} k$ for some scalar $\lambda_{ij}$. In fact, up to scalar multiples, the group generated by $\{1,i,j,k\}$ must be the Klein 4-group. By classification of small groups, the smallest candidate must be the Quaternion group of order 8. The products can't actually be the Klein 4-group because rotations aren't commutative. The quaternion group works, so we are done.

Finally, what is $ij$? Well, for geometric reasons it must be $\lambda_{ij} k$ for some scalar $\lambda_{ij}$. Moreover, we have $(iji^{-1})^2 = ij^2i^{-1}=-1$; so $\lambda_{ij}=\pm 1$. By geometric considerations also, $ji = \lambda_{ji}k$, and we can likewise conclude that ${ji} = \pm k$. Finally, we conclude that $i$ and $j$ anti-commute by starting with $(ij)^2 = -1$ and rearranging to get $ij = -(ij)^{-1}=-j^{-1}i^{-1}=-ji$.

[Removed an opinionated remark that's perhaps non-mathematical. If someone who's seen it thinks it's appropriate to include it here, then I might do that].

[Removed a subjective remark that's perhaps non-mathematical. If someone who's seen it thinks it's appropriate to include it here, then I might do that].

Subjective criticism of Clifford algebra

Feel free to tell me in the comments whether I should remove what I'm about to write. It may perhaps be opinion and not mathematics.

The following is entirely subjective. My issue with Clifford algebra is that a general element of a Clifford algebra lacks meaning. Normally, you restrict your attention to the even subalgebra or the odd subalgebra. The even subalgebra represents rotation-like transformations, whereas the odd subalgebra represents reflection-like transformation (general roto-reflections). A linear combination of these two "subalgebras" is often meaningless, unless you go to higher dimensions, where they are revealed to be rotations in one dimension higher; but then the mystery simply repeats itself at a higher dimension. It's also not clear why an element and its negation represents the same transformation, unless you approach it as I have: as a form of homogeneous or projective representation.

Actually, my discomfort is with the idea of formally adding a scalar to a vector to a bivector. But I'll stop here. The approach I'm taking has the advantage that every element has some meaning as a transformation, instead of some elements being vectors, and some being bivectors, and some apparently not meaning anything. But it's also possible that understanding all elements as transformations causes one to miss out on exterior algebra.

[Removed an opinionated remark that's perhaps non-mathematical. If someone who's seen it thinks it's appropriate to include it here, then I might do that].