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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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    $\begingroup$ @Pedro: Presumably this is akin to a tease. We know very little about homotopy groups of spheres, in particular how to compute them. My impression is the intention of the post is to trigger an ambitious young homotopy theorist to crank out some computation or theory, and answer the question. $\endgroup$
    – Ryan Budney
    Dec 30, 2022 at 8:59

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