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I know that some $ S^n $ s have group structures, whereas others do not.

For example,

$ S^0 = \{\pm 1\} \cong \mathbb{Z}/2\mathbb{Z}$,

$ S^1 = \mathbb{R}/\mathbb{Z} $,

$ S^3 = \{a+bi+cj+dk:a^2+b^2+c^2+d^2=1\} $, a multiplicative subgroup of the real quaternion algebra.

On the other hand, $ S^{2n} $ with $ n \geq 1 $ cannot have a group structure. Proposition 2.29 of A. Hatcher's Algebraic Topology (on page 135) states that $ \mathbb{Z}/2\mathbb{Z} $ is the only nontrivial group that can act freely on $ S^{2n} $. But if $ S^{2n} $ itself was a group, then it acts freely on itself.

I want to know whether there is a complete classification theorem about for which $ n $ does $ S^n $ admit a group structure (not necessarily a Lie group structure). Thanks in advance!

syn3449
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  • https://math.stackexchange.com/questions/12453/is-there-an-easy-way-to-show-which-spheres-can-be-lie-groups – Nick L Feb 20 '18 at 23:05
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    You say you don't require a Lie group structure, but do you still want a topological group structure? Because if not, then since all spheres have the same cardinality (apart from the zero dimensional one) they all have at least one group structure. – Arnaud D. Feb 20 '18 at 23:21
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    @ArnaudD. I think the op requires that the topology be compatible with group operations, i.e. a topological group structure. – Raman Apr 13 '22 at 10:55

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