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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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