Show that the group \(\mathbb{Z}_p \times \mathbb{Z}_p\) with \(p\) prime cannot act freely on any sphere \(S^n\).
Solution
Let \[ G=\mathbb Z_p\times\mathbb Z_p. \] Suppose first \(n>1\) and that \(G\) acts freely on \(S^n\). Then \[ S^n\longrightarrow M=S^n/G \] is the universal covering of a closed \(n\)-manifold, with \[ \pi_1(M)=G, \qquad \pi_i(M)=0\quad(1<i<n), \qquad \pi_n(M)\cong\mathbb Z. \] Attach one \((n+1)\)-cell along a generator of \(\pi_n(M)\), then cells of dimensions at least \(n+2\) to kill all remaining higher homotopy groups. This produces a \(K(G,1)\).
Up through dimension \(n+1\) this CW complex has exactly one new \((n+1)\)-cell beyond the \(n\)-dimensional manifold \(M\). With \(\mathbb F_p\)-coefficients the cochain group in degree \(n+1\) is therefore one-dimensional. Consequently \[ \dim_{\mathbb F_p}H^{n+1}(K(G,1);\mathbb F_p)\le1. \]
But \[ K(G,1)\simeq K(\mathbb Z_p,1)\times K(\mathbb Z_p,1). \] For \(p=2\), \[ H^*(K(\mathbb Z_2,1);\mathbb F_2)\cong\mathbb F_2[u],\quad |u|=1, \] and for odd \(p\), \[ H^*(K(\mathbb Z_p,1);\mathbb F_p) \cong\Lambda(u)\otimes\mathbb F_p[v], \quad |u|=1,\ |v|=2. \] In either case the Künneth formula shows that the degree-\(d\) cohomology of the product has dimension \(d+1\). In particular \[ \dim H^{n+1}(K(G,1);\mathbb F_p)=n+2>1, \] a contradiction.
If \(n=1\), any finite group acting freely on \(S^1\) is cyclic: an orientation-reversing homeomorphism of \(S^1\) has a fixed point, so a finite free action is orientation-preserving and conjugate to a rotation action. Since \(\mathbb Z_p\times\mathbb Z_p\) is not cyclic, it cannot act freely on \(S^1\) either.
Thus \[ \boxed{\mathbb Z_p\times\mathbb Z_p\text{ acts freely on no sphere}.} \]