Finite abelian groups that can act freely on \((S^{2n})^k\) (Q1091643)

Let \(X\) be a finite CW complex of the homotopy type of \(S^{2n}\times\cdots\times S^{2n}\) \((k\) copies) and let \(G\) be a finite abelian group acting freely and cellularly on \(X\). The author observes that \(G\) must be a 2-group, that is, \(G\cong G_j\) where each \(G_j\) is cyclic of order \(2^{m(j)}\). The main result is that the orders of these cyclic groups must satisfy the inequality: \(2k\geq \sum |G_j|\). The key to the proof is a clever use of Lefschetz fixed point theorem to establish that since \(G\) acts freely, the induced homomorphisms \(g^*\colon H^{2n}(X; \mathbb{Q})\to H^{2n}(X; \mathbb{Q})\), for \(g\in G\), give a faithful representation of \(G\) on \(H^{2n}(X; \mathbb{Q})\simeq \mathrm{GL}(k; \mathbb{Q})\). The bound on the orders of the \(G_j\) then is proved by representation theory. The author notes that the bound is sharp in the sense that, given \(k=\sum 2^{m(j)-1}\), the group \(G=\prod G_j\) (where \(G_j\) is cyclic of order \(2^{m(j)})\) acts freely and cellularly on \(S^{2n}\times\cdots\times S^{2n}\) \((k\) copies).