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Say we have a periodicity generator $e \in H^k(BG)$. I can show that we then have a $(k-1)$-dimensional $G$-complex $X$ with free $G$-action. It's also not that difficult to see that it has trivial $G$-action on $H^{k-1}(BG)$ with $e(X)=e$. However, my mentor claims that $X$'s oriented $G$-homotopy type must be uniquely determined by the periodicity generator $e$. I just can't see how.

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Take $EG\to BG$ to be the universal $G$ principal bundle of $G$. Fix $X$ to be an homotopy representation of $G$ with free action of dimension $n-1$. Considerer the fiber bundle $p : EG\times_G X \to BG$. This one is a spherical bundle. If $G$ acts on $H^{n-1} (X,\mathbf{Z})$ then we have an exact sequence $$\ldots \to H^k (BG) \to H^{k+n} (BG) \to H^{k+n}(EG\times_G X)\ldots \to$$ where the first map is noted $e_X$ and the second one is the one induced by $p$. The map $e_X$ is just the multiplication by the Euler class $e(X)\in H^{n}(BG)$. Defining this Euler class requires an orientation of our sphere bundle, and this is given by an element of $H^{n-1}(X)$ that generates $H^{n-1}(X)$. The projection $EG\times_G X \to X/G$ is a fibration with contractible fibres all equal to $EG$ so that you can show that it is an homotopy equivalence. Now, as $X/G$ is of dimension $n-1$, you get that $EG\times_G$'s cohomology vanishes in degree $k\geq n$, and then $e_X$ is an isomorphism for all $k>0$. Finally, as ($\mathbf{Z}$-valued) cohomologies of $BG$ and $G$ are the same, you see that $G$ has periodic cohomology, and that $e$ is a periodicity generator.

What you are stating, plus yours professor's statement is the reciprocal of what preceeds, and was proved by R.G. Swan in Periodic Resolutions for Finite Groups, Annals of Mathematics, Second Series, Vol. 72, No. 2 (Sep., 1960), pp. 267-291, and it is avalaible for free in online reading. It is its theorems A and A', and your professor's statement is consequence of results proved in the article, namely, the generalization of Schanuel's theorem.

Olórin
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