at.algebraic topology – \$pi_{2n-1}(operatorname{SO}(2n))\$ element represents the tangent bundle \$TS^{2n}\$, not torsion and indivisible?

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 $$malpha$$ 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 $$alpha=kbeta$$.

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

transactions – Why UTXO is indivisible

I am new to understanding Bitcoin, but just read this:

If a UTXO is greater than the desired value of a transaction, it must always be consumed in its entirety and the change must be generated in the transaction. In other words, if you have a UTXO with a value of 20 bitcoins and you only pay for one bitcoin, your transaction must consume the entire UTXO code of 20 bitcoins and generate two outputs : one paying 1 bitcoin to the desired recipient and the other paying 19 bitcoin in modification. Back to your wallet. Due to the indivisible nature of the transaction results, most bitcoin transactions will have to generate changes.

Not knowing much, I do not understand why you can not simply divide the bitcoin arbitrarily as an integer. I saw these:

You wonder if this decision was made for an important reason or if it's just "it's like that because it's like that".