TITLE: Topological structure of SO(n) as a product
QUESTION [7 upvotes]: I’m interested in the question for which $n$ the special orthogonal group is homeomorphic to the product
$$ \mathrm{SO}(n) \approx S^{n-1} \times \mathrm{SO}(n-1). $$
Allen Hatcher [1, p. 293 f.] claims (?) that this is true for $n \in \{ 2, 4, 8 \}$ and wrong for all other values (although I’m not sure what is meant by “twisted product”).
Does anyone have a reference where this is done in more detail? Maybe I just didn’t have the right keywords for a proper search.
[1] Algebraic Topology, downloadable at https://www.math.cornell.edu/~hatcher/AT/ATpage.html

REPLY [18 votes]: Another line of reasoning: If $SO(n)$ were homeomorphic to the product $S^{n-1}\times SO(n-1)$ then $S^{n-1}$ would be a retract (not deformation retract!) of $SO(n)$, hence $S^{n-1}$ would be an H-space since a retract of an H-space is an H-space, assuming the retract contains the identity element, which we can arrange here by a suitable choice of retraction.  Then Adams' theorem restricts $n$ to $2,4,8$.