$\pi_{2n-1}(\operatorname{SO}(2n))$ element represents the tangent bundle $TS^{2n}$, not torsion and indivisible for $n>1$?

Question

Question: Is the element $\alpha$ in $\pi_{2n-1}(\operatorname{SO}(2n))$ representing the tangent bundle $TS^{2n}$ of the sphere $S^{2n}$ indivisible and not torsion?

My understanding so far —

An $\operatorname{SO}(2n)$ bundle over $S^{2n}$ corresponds to an element in $\pi_{2n}\operatorname{BSO}(2n) =\pi_{2n-1}\operatorname{SO}(2n)$.

Not torsion: There does not exist any integer $m > 0$ such that $m\alpha$ is a trivial element.

Indivisible: There does not exist any integer $k > 1$ and any element $\beta$ in $\pi_{2n-1}\operatorname{SO}(2n)$ such that $\alpha=k\beta$.

Ref: Mimura, Toda: Topology of Lie groups. Chapter IV Corollary 6.14.

p.s. I am asking $n>1$ so if you vote down for $n=1$, you may reconsider your vote... Lol

Answer 1

For $n=1$, the answer to your question is negative, as explained by Gregory Arone in the comments.

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In the cases $n\neq 1,2,4$, there is the following easy argument:

The long exact sequence of the fibration $S^{2n-1}\to BO(2n-1)\to BO(2n)$ induces an exact sequence $$\pi_{2n}(BSO(2n))\xrightarrow{e} \mathbb{Z}\xrightarrow{[TS^{2n-1}]}\pi_{2n-1}(BSO(2n-1)),$$ where $e$ is the evaluation of the Euler class. As the order of $[TS^{2n-1}]\in \pi_{2n-1}(BSO(2n-1))$ is $2$ unless $n=1,2,4$, the image of $e$ is $2\mathbb{Z}$. As $\chi(S^{2n})=2$, we have $e([TS^{2n}])=2$, so $[TS^{2n}]$ must be nontorsion and indivisible as long as $n\neq 1,2,4$.

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Let me also mention the following convenient description of $\pi_{2n}(BO(2n))$ (this can be used to settle the cases $n=2,4$):

Combining the Euler class with the inclusion $SO(2n)\subset SO$, we have a morphism $$ (e,s)\colon \pi_{2n}(BO(2n))\rightarrow \mathbb{Z}\oplus \pi_{2n}(BO)=\mathbb{Z}\oplus\begin{cases} \mathbb{Z} & \text{ if }n\equiv 0,2\text{ (mod) } 4\\ \mathbb{Z}/2 & \text{ if }n\equiv 1\text{ (mod) } 4\\ 0 & \text{ if }n\equiv 3\text{ (mod) } 4 \end{cases}. $$ From Kervaire's Some nonstable homotopy groups of Lie groups, one can deduce that as long as $n\neq 1$ this morphism is injective with image $$ \begin{cases} 2\mathbb{Z}\oplus \mathbb{Z} & \text{ if }n\equiv 0\text{ (mod) } 4,n\neq 2,4\\ \{(k,l)\in\mathbb{Z}^2|k+l\text{ even} \} & \text{ if }n=2,4\\ 2\mathbb{Z}\oplus\mathbb{Z}/2 & \text{ if }n\equiv 1\text{ (mod) } 4\\ 2\mathbb{Z}\oplus 0 & \text{ if }n\equiv 3\text{ (mod) } 4. \end{cases} $$ As $S^{2n}$ is stably parallelisable, the image of $[TS^{2n}]$ under this morphism is $(2,0)$, so this class is indivisible and nontorsion.