I was asked to show if $$\mathbb{S}^{n-1}\rightarrow \mathbb{S}^{2n-1}\rightarrow \mathbb{S}^{n}$$ holds. Then $n=1,2$ or divisible by $4$. The $n=1,2$ cases in the converse are verfied. I am wondering how to prove the second claim.
The thought is to use the relationship $$\pi_{j}(\mathbb{S}^{n})=\pi_{j}(\mathbb{S}^{2n-1})\oplus \pi_{j-1}(\mathbb{S}^{n-1})$$as the exact sequence splits. We can use Serre's result which asserts $\pi_{i}(\mathbb{S}^{j})$ are all finite except in the case $i=j,i=4k-1,j=2k$. If we let $j=2n-1$, then we have: $$\pi_{2n-1}(\mathbb{S}^{n})=\pi_{2n-1}(\mathbb{S}^{2n-1})\oplus \pi_{2n-2}(\mathbb{S}^{n-1})$$Since the middle group is $\mathbb{Z}$, this only holds if $n$ is even. But I cannot prove $n$ must be divisible by $4$.