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Let $G$ be a finite group. The theorem that the Burnside ring $A(G)$ is isomorphic to the zeroth stable stem $\pi^{G}_0(S)$ is usually said to originate from Segal. I search for a reference of a proof this theorem, which is not obscured by tom Diecks generalisation to compact Lie groups. Who knows a good reference? I can not even find the right paper of Segal. Where did the original proof appear?

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This is from Segal's paper

Equivariant stable homotopy theory. Actes du Congrès International des Mathématiciens (Nice, 1970), Tome 2, pp. 59--63. Gauthier-Villars, Paris, 1971.

MR0423340 (54 #11319)

where he states that the equivariant stable cohomotopy of $S$ is $\pi_G^{-\ast}(S)=\bigoplus_K\pi_{\ast}^s(B(N_G(K)/K)^+)$. In particular, $\pi_G^0(S)$ is the Burnside ring.

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Thanks for your answer! I got a copy from the library. Do you know of any alternative account, written may be in a more teaching style? –  user2146 Jan 22 '10 at 10:14
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Oh, sorry, I missed this. The only other reference I know for a proof of this is Lewis-May-McClure from LMN 1213, Chapter V, § 2. (This is available for download from May's "old books page" if you're interested.) Though they treat the general case, you may find it easy to extract those portions of the discussion that are relevant for your purposes. –  Clark Barwick Feb 8 '10 at 18:53
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Here's a conceptual answer, which can be filled in to give a proof. First, go back to the non-equivariant setting: why is $\pi_0(S) = \lim \pi_N(S^N) \cong {\mathbb Z}$? because one can use transversality arguments (originally due to Pontryagin) to show that group is the same as the cobordism group of ``oriented'' zero-dimensional manifolds, which is then obviously the integers. Similarly, the cobordism of oriented zero-dimensional $G$-manifolds is isomorphic to the Burnside ring, so the result you seek follows once you can establish some transversality for maps $S^V \to S^V$. (These transversality results are notoriously complicated and not always valid in the equivariant setting, but this one doesn't present much trouble).

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