Topological structure of SO(n) as a product 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
 A: This is another formulation of the celebrated Hopf invariant theorem of Adams. If the decomposition holds, then the fibration
$$SO(n−1)\to SO(n)\to S^{n−1}$$
must admit a section $S^{n−1}\to SO(n)$ which in particular is nontrivial in homology. Now, compose this with the unstable J-homomorphism $SO(n)\to \Omega^n S^n$. The composition still is nontrivial in $\mathbb{Z}/2$-homology. The outcome then is a spherical class in $H_{n-1}(\Omega^nS^n;\mathbb{Z}/2)$ that is coming from the image of J-homomorphism. But, it is known that the only spherical classes in these spaces, coming from the image of J-homomorphism,  are the Hopf invariant one elements, that is this can happen if and only if the adjoint map $S^{2n-1}\to S^n$ is detected by the mod 2 Hopf invariant. This is possible only in the above dimensions according to Adams. 
A: If this was the case, then $\pi_{n-2}(SO(n))\cong \pi_{n-2}(SO(n-1))$. The first group is in the stable range and hence one of $\mathbb{Z},\mathbb{Z}/2$ or 0; in particular every surjective endomorphism of it is an isomorphism. Now the orthonormal frame bundle of $S^{n-1}$ is a fiber sequence $SO(n-1)\to SO(n) \to S^{n-1}$ which gives a long exact sequence
$$
\pi_{n-1}(SO(n))\to\pi_{n-1}(S^{n-1})\to \pi_{n-2}(SO(n-1))\to \pi_{n-2}(SO(n))\to\pi_{n-2}(S^{n-1}).
$$
The last group is zero, so the third arrow is a surjection; by what was said before, it is an isomorphism. Hence the first map is surjective, and a preimage of the canonical generator gives a section of the frame bundle, in other words, a trivialization of the tangent bundle of $S^{n-1}$. But it is well-known (see user51223's answer) that this is only possible in the dimensions you have listed above.
A: 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 $2,4,8$.
A: I think this is not an answer but a precisation. Assume that the statement is true for some $n$. 
Then S^{n-1}\times SO(n-1)has a Lie group strcture and in particular a parallelizable tangent bundle. This implies that the tangent bundle of S^{n-1} is parallelizableand this is true if and only if n=2,4,8 (the original reference is //projecteuclid.org/euclid.bams/1183522319). Edit: As explained below (by Bertram Arnold), this implies that the sequence 
$SO(n-1)\to SO(n)\to S^{n-1}$
splits, and this split give rise to a section in the frame bundle of $S^{n-1}$, i.e it gives a trivialization of the tangential bundle of $S^{n-1}$. This is true if and only if $n=2,4,8$ (the original reference is https://projecteuclid.org/euclid.bams/1183522319).
Now what about the group structure?
a) If $n=2$ then $S^{1}\times SO(1)\cong S^{1}$
b) If $n=3$, then $SO(4)\cong S^{3}\times SO(3)$, in particular SO(4) is diffeomophic to a product of two Lie groups  but it isn't a Lie groups isomorphism (see the comments below). (The Lie group structure on $S^{3}$ is the one give by the quaternions with norm $1$)
c)  If $n=8$, then $SO(8)\cong S^{7}\times SO(7)$, but it is not longer a product of Lie group . $S^{7}$ may be viewed as the set of octonions with norm 1, it is in particular a quasi Lie group (or non associative Lie group).
