Signed permutations in $ \operatorname{SO}(n) $ and normalizing an extraspecial group

Question

$\DeclareMathOperator\SO{SO}\DeclareMathOperator\O{O}\DeclareMathOperator\SU{SU}\DeclareMathOperator\Lift{Lift}$The subgroup of $ \SO(n) $ of signed permutations has order $ n!2^{n-1} $. I will call this group $ W_n $.

I think that $ W_n $ fails to be maximal if and only if $ n=2^k $ is a power of $ 2 $. Indeed it seems to me that $ W_{2^k} $ normalizes a certain extraspecial subgroup $ E \subset \SO(2^k) $ of order $ 2^{2k+1} $. I think the full normalizer of $ E $ is generated by the signed permutations together with tensor products involving $$ \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$ Is this true?

Context:

$W_3 \cong S_4$ is isomorphic to the symmetric group on 4 letters and $ W_3 $ is maximal.

$W_2 \cong C_4$ is the cyclic group of order 4. So $ W_2 $ is not maximal.

And for $ W_4 $ we have a chain of strict containments $$ W_4 \subsetneq \Lift(S_4 \times S_4) \subsetneq \SO(4) $$ where $\Lift$ denotes the Lift through the double cover $ \SO(4) \to \SO(3) \times \SO(3) $. So $ W_4 $ is also not maximal.

Answer 1

This is more an extended comment than an answer, but it wouldn't fit in the comments section.

There seems to be some confusion in the original question as well as the comments, between the group of monomial (or signed permutation) matrices in $SO(n)$ and the Weyl group. The monomial matrices form a group of shape $2^{n-1}S_n$. The extension does not split for $n$ even, but it has a subgroup of index two that does split, of shape $2^{n-1}A_n$. For the Weyl group, on the other hand, you need to choose a maximal torus, which is a product of $\lfloor\frac{n}{2}\rfloor$ copies of $SO(2)$, and which then pairs up the coordinates. This is why $n$ even and $n$ odd behave differently. The Weyl group of $SO(2n)$ is $D_n$, while the Weyl group of $SO(2n+1)$ is $B_n$.

The full normaliser of your extraspecial group $E$ of order $2^{1+2k}$ in $SO(2^k)$ is a (usually non-split) group of shape $2^{1+2k}O^+_{2k}(2)$, where $O^+_{2k}(2)$ is the appropriate orthogonal group in characteristic two. See Griess, "Automorphisms of extra special groups and nonvanishing degree 2 cohomology", especially Theorem 5. However, this doesn't contain your monomial group for $k$ large, because $2^{2^k-1}.(2^k)!$ grows much more rapidly than the size of this group. This probably means that the "only if" part of your statement is wrong.

Finally, if an appropriate amendment of your statement is true (I haven't checked it), it should be possible to deduce it from Collins, "On finite subgroups of the classical groups" without too much trouble. See also Weisfeiler, "Post-classification version of Jordan's theorem on finite linear groups" for related statements.

Answer 2

An example rotation matrix between integral octonions is this matrix:

$$\frac{1}{2}\begin{bmatrix}1 & 1 &1&0&1&0&0&0\\ -1&1&1&0&-1&0&0&0\\ -1&-1&1&0&1&0&0&0\\ 0&0&0&1&0&1&-1&1\\ -1&1&-1&0&1&0&0&0\\ 0&0&0&-1&0&1&1&1\\ 0&0&0&1&0&-1&1&1\\ 0&0&0&-1&0&-1&-1&1\end{bmatrix}$$

Because we also allow any permutation, one choice of the (half-)integral octonions is not invariant under the entire rotation group. The union of all $7$ variants of (half-)integral octonions is: if you take any set with $4$ distinct basis octonions, their sum times $0.5$ appears in one of the $7$ integral octonion algebras.

Edit: Unfortunately this idea of me doesn't work. By conjugating with a permutation you can get a multiplication map for unit integral octonions from each of the seven integral octonion algebra's. Since that is not a multiplicatively closed set, we will obtain all multiplications with unit octonions.