I can calculate the derivation of the octonions and I clearly find the 14 generators that form the algebra of $G\_2$. However, when I do the same calculation for the quaternions, I end up with the three generators of $SO(3)$, which basically tells me that I can rotate the set of imaginary units anyway I like.


Intuitively, I don't understand, why it is not possible for the octonions to be rotated in the same way with an arbitrary rotation of $SO(7)$. Instead, the calculations show, that the possible transformations are restricted to the $G\_2$ subgroup.


Is there way to understand this geometrically or algebraically? Is it related to the non-associative property of the octonions?



Why is the automorphism group of the octonions $G_2$ instead of $SO(7)$
The quaternions are generated by any *two* imaginary elements $x$ and $y$ that are orthonormal, i.e., $\bigl(1,\, x,\, y,\, xy\bigr)$ is an orthonormal basis of the quaternions. Moreover, the multiplication table using such a pair does not depend on which pair you choose. That's why the automorphism group of the quaternions acts transitively on orthonormal pairs in the imaginary quaternions.


Meanwhile, the octonions are generated algebraically by any *three* imaginary elements, say, $x$, $y$, and $z$ that are orthonormal *and* $z$ is perpendicular to $xy$. In fact, $\bigl(\,1,\, x,\, y,\, xy,\, z,\, zx,\, zy,\, z(xy)\,\bigr)$ is an orthonormal basis for the octonions. Moreover, as was shown by Dickson, if $x'$, $y'$, $z'$ are another orthonormal triple of imaginary octonions such that $z'$ is perpendicular to $x'y'$, there is a unique automorphism of the octonions that carries $(x,y,z)$ to $(x',y',z')$.


Note that the set of orthonormal triples $(x,y,z)$ in the imaginary octonions is the Stiefel manifold $V\_{7,3}$, which has dimension $6+ 5+ 4 = 15$, and the single extra relation $z\cdot xy = 0$ cuts out a submanifold of dimension $14$. Hence the automorphism group of the octonions is a Lie group of dimension $14$.


Note that, any automorphism of the octonions that fixes three such elements is the identity. Thus, $\mathrm{SO}(7)$ (which has dimension $21$) is too large to be the automorphism group of the octonions, not the least because it acts transitively on the set of oriented orthonormal bases of the imaginary octonions.


(Sadly, when $x$, $y$, $xy$, and $z$ are imaginary orthogonal octonions, the defining property $uv\cdot uv = (u\cdot u)(v\cdot v)$ turns out to imply that $(xy)z = -x(yz)$, so the octonions are not associative. However, this is not really the reason that $\mathrm{SO}(7)$ is not the automorphisms of the octonions.)


