Let $p$ be an odd prime and $n>1$.
I want to prove that $\pi_i(S^n)$ has no $p$-torsion for $i<n+2p-3$.
For odd $n$ this is Proposition 6.26 in McCleary's book (p.206). He mentions afterward the result for even-dimensional spheres can be obtained by 'similar arguments'.
His argument for odd $n$ went like this:
- There is a fibration $K(\Bbb Z,n-1)\to X\to S^n$ and $X$ has the same homotopy groups as $S^n$ in degrees $\geq n+1$ and zero in degrees $\leq n$.
- It suffices to determine the connectivity of $H^*(X;\Bbb Z_p)$ because then we can apply the Hurewicz-Serre theorem which says: $$H_i(X;\Bbb Z_p)=0 \text{ for } 0<i<m \implies \pi_i(X)\otimes \Bbb Z_p =0 \text{ for } i<m$$
- The $\Bbb Z_p$ cohomology of the fiber is known: It's a free graded-commutative algebra on certain elements, and starts with $H^{n-1}(K(\Bbb Z,n-1);\Bbb Z_p)=\Bbb Z_p\cdot\iota_{n-1}$. Since $n$ is odd, all the powers $(\iota_{n-1})^k$ exist as well. The class of next highest degree is $P^1(\iota_{n-1})$.
- $\iota_{n-1}$ transgresses to a generator $u\in H^n(S^n;\Bbb Z_p)$, hence the derivation property implies $d((\iota_{n-1})^k)=u\otimes (\iota_{n-1})^{k-1}$ for $k<p$, so none of those products persist to $E_\infty$.
- This implies that the class of least degree which persists to $E_\infty$ is $P^1(\iota_{n-1})\in H^{n-1+2(p-1)}(X;\Bbb Z_p)$.
- Thus $H^*(X;\Bbb Z_p)$ starts in degree $n-1+2(p-1)=n+2p-3$ where it is $\Bbb Z_p$. Now the Hurewicz-Serre theorem implies the claim.
It is not at all clear to me how to transfer this argument to the case that $n$ is even.
If $n$ is even, there are no powers of $\iota_{n-1}$ by graded commutativity, so the product $u\otimes \iota_{n-1}$ does not get killed. And actually this element comes from the fact that $\pi_{2n-1}(S^n) = \Bbb Z$, so there is no way it could be killed. There seems to be no way of applying Hurewicz-Serre because the conncetivity of $H^*(X;\Bbb Z_p)$ is terrible. Can this situation be remedied?
Remarks:
The fibration is obtained by pulling back the pathspace fibration $K(\Bbb Z, n-1)\to P \to K(\Bbb Z,n)$ along a map $S^n\to K(\Bbb Z,n)$ which is a $\pi_n$-isomorphism.
I also tried using another space instead of $X$ - one which has the same homotopy groups as $S^n$ in degrees $\geq 2n$ and zero below, but in that case we don't know much about fiber or base space.