If we localize at 2, is $S^n$ an H-space? It is well known that the sphere $S^n$ is an H-space iff $n = 0, 1, 3, 7$. This has been proved by Adams in On the non-existence of elements of Hopf invariant one, but I have never read it and I don't know how the proof goes. However, this is a result for the "integral" spheres, but things change if we start inverting primes.
For example, if we just invert 2, then every odd sphere is an H-space. In fact, Serre proved in Groupes d'homotopie et classes de groupes abéliens that $\Omega S^{2n} \simeq S^{2n-1} \times \Omega S^{4n-1}$ away from 2 and hence $S^{2n-1}$ is an H-space since it is a retract of an H-space.
On the other side, even if we localize at an odd prime, no even sphere can be an H-space. In fact $H(S^{2n}; R) = R\langle 1, i_{2n}\rangle$ with $\Delta i_{2n} = i_{2n} \otimes 1 + 1 \otimes i_{2n}$.  If it would be an H-space, then $0 = \Delta i_{2n}^2 = (i_{2n} \otimes 1 + 1 \otimes i_{2n})^2 = 2 \ i_{2n} \otimes i_{2n}$ and this cannot happen unless $2 = 0$ in $R$.
Hence, these arguments solve the problem for all spheres at all odd primes. But both these arguments don't work at 2.
What happens at 2? If we localize at 2, can a sphere be an H-space, other than for $n = 0, 1, 3, 7$?

*

*I think that, at least in theory, odd spheres could have two H-space structures, one at 2 and one away from 2, that don't "match up" integrally; and even spheres could have an H-space structure at 2, because the "integral failure" already happens at odd primes.

*Nevertheless, in practice, I guess that $S^n$ cannot be an H-space unless $n = 0, 1, 3, 7$, even if we localize at 2. My feeling is that the problem is exactly at 2. Maybe Adams proved exactly that this cannot happen, but since I don't know the details of the paper...

Thanks in advance!
 A: We work with the mod $2$ Steenrod algebra $\mathcal{A}$ and take $\mathbb{Z}/2$ coefficients in all cohomology groups.
Start with the Adem relation
$$Sq(2^k)Sq(2^{k+1}l)=Sq(2^{k+1}l+2^k)+\text{decomposables}.$$
By induction on degree this identity yields:

The set $\{Sq^{2^k}\mid k\geq0\}$ generates the mod $2$ Steenrod algebra $\mathcal{A}$.

On the other hand, if we take a generator $u\in H^1\mathbb{R}P^\infty$ we have $Sq^{2^k}u^{2^{k}}=u^{2^{k+1}}$ and $Sq^ju^{2^k}=0$ whenever $0<j<2^k$. Thus:

$Sq^{2^k}$ is indecomposable in $\mathcal{A}$ for each $k\geq0$.

So suppose $S^n$ admits an H-multiplication. The Hopf construction yields a map $\varphi:S^{2n+1}\rightarrow S^{n+1}$ whose mapping cone $C_\varphi=S^{n+1}\cup_\varphi e^{2n+2}$ has
$$H^*C_\varphi=\mathbb{Z}_2[x]/(x^3),\qquad |x|=n+1.$$
Thus
$$Sq^{n+1}x=x^2\neq 0.$$
From the first fact above we immediately see that $n+1$ is a power of $2$. Otherwise $Sq^{n+1}$ decomposes as a polynomial in Steenrod operations of lower degrees. Since $C_\varphi$ has no cohomology in degrees $n+1,\dots,2n-1$, this would result in $Sq^{n+1}$ annihilating $x$.
Now all of these ideas have really been 2-local. The conclusion suffers no loss if we take as input a 2-local sphere. All the cohomology groups factor through the 2-local category, and the facts about the Hopf construction hold in much greater generality.

If the 2-local sphere $S^n$ admits an H-multiplication, then $n+1$ is a power of $2$.

The difficult part is to now discount the cases $n+1=2^k$ with $k>3$. The second fact stated above shows that more intricate methods must be employed. Adams's observation is that whenever $k> 3$, then the operation $Sq^{2^k}$ can in fact be decomposed outside of $\mathcal{A}$, at least in a certain sense.
More precisely Adams constructs secondary operations $\Phi_{i,j}$ for $0\leq i\leq j$, $j\neq i+1$, of degrees $2^i+2^j-1$, and demonstrates that the formula
$$Sq^{2^{k+1}}x=\sum_{i,j<k} a_{i,j,k}\Phi_{i,j}(x)+\text{indeterminacy}$$
holds on cohomology classes $x$ satisfying $Sq^{2^r}x=0$ for $0\leq r\leq 2^k$. Here $k\geq3$ and the coefficients $a_{i,j,k}$ are certain primary cohomology operations.
As an example, if $v\in H^2\mathbb{C}P^\infty$ is a generator, then $\Phi_{0,i}$ is defined on $v^{2^j}$ for $i\leq j$ and evaluates as $\Phi_{0,j}(v^{2^j})=v^{3\cdot 2^{j-1}}$ with no indeterminacy, and $\Phi_{0,i}(v^{2^j})=0$ with no indeterminacy for $i<j$.
Returning to our putative 2-cell complex $C_\varphi$ and assuming $n+1=2^k$, $k>3$, the secondary operations $\Phi_{i,j}$ are defined on $x$. There is no trouble in computing the indeterminacy in the above formula, and it vanishes for reasons of degree. Thus Adams has succeeded in factoring the action of $Sq^{n+1}$ on this complex through trivial cohomology groups. In particular $Sq^{n+1}x=0$, which is at odds with the construction of $C_\varphi$. The conclusion is that the 2-local complex $C_\varphi$ cannot exist.

Adams: If $n\neq 0,1,3,7$, then the 2-local sphere $S^n$ cannot admit an H-space structure.

