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Someone asked me if all finite abelian groups arise as homotopy groups of spheres. I strongly doubted it, and I bet ten bucks that $\mathbb{Z}_5$ is not $\pi_k(S^n)$ for any $n,k$. But I don't know how to prove it's not.

Which finite abelian groups are known to not arise as homotopy groups of spheres?

I conjecture that $\mathbb{Z}_5$ is the smallest one. From some tables we can see that all smaller groups do actually arise:

$$ \begin{array}{ccl} \pi_1(S^2) &\cong & 1 \\ \pi_4(S^3) &\cong& \mathbb{Z}_2 \\ \pi_9(S^3) &\cong& \mathbb{Z}_3 \\ \pi_8(S^4) &\cong& \mathbb{Z}_2 \times \mathbb{Z}_2 \\ \pi_{122}(S^{62}) &\cong& \mathbb{Z}_4. \end{array} $$

In fact, I conjecture that for no odd prime $p \gt 3$ is $\mathbb{Z}_p$ isomorphic to $\pi_k(S^n)$ for any $n,k$.

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    $\begingroup$ I think every cyclic group appears as a subgroup of a homotopy group of spheres. Indeed, I think this is already true for the image of j. But I think homotopy groups of spheres are typically groups are “small” unless n-k=0 mod 4, in which case the image of j is already quite big, so it does seem likely that most finite groups will not appear. $\endgroup$
    – Theo Johnson-Freyd
    Nov 26, 2020 at 23:09
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    $\begingroup$ The cyclic group of order 4 appears in the table <en.wikipedia.org/wiki/…> of stable homotopy groups of spheres reproduced in Wikipedia's entry for "Homotopy groups of spheres". So it seems that this group appears as $\pi_{n+60}(S^n)$ for $n$ sufficiently large. I do not see a 5-element group tabulated anywhere on that page. $\endgroup$
    – Noam D. Elkies
    Nov 26, 2020 at 23:15
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    $\begingroup$ Are there heuristics (a la Cohen-Lenstra) for p-part of the homotopy groups of spheres mod the image of J? Or maybe in the stable case? $\endgroup$
    – Noah Snyder
    Nov 26, 2020 at 23:57
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    $\begingroup$ I can't give you any concrete answer, but my inclination is to say that it seems highly likely that you're correct. The 2-primary torsion forms a real thicket, whereas there can be no p-torsion until you are at least at $\pi_{n+2p-3}(S^n)$. This means it's not clear to me whether there are infinitely many different groups of odd order appearing. (But the unstable part is really outside my wheelhouse...) $\endgroup$
    – Tyler Lawson
    Nov 27, 2020 at 5:58
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    $\begingroup$ @CalicoJackRackham I think you mean "only two normal covering spaces" rather than "covering space". After all non-cyclic simple groups have a lot of subgroups. But since the question is only interesting for higher homotopy groups (which are always abelian) the relevance of your comment is not clear to me. $\endgroup$
    – Denis Nardin
    Dec 11, 2020 at 15:22

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