Show that if \(\tilde{H}^n(X; \mathbb{Q})\) and \(\tilde{H}^n(X; \mathbb{Z}_p)\) are zero for all \(n\) and all primes \(p\), then \(\tilde{H}_n(X; \mathbb{Z}) = 0\) for all \(n\), and hence \(\tilde{H}^n(X; G) = 0\) for all \(G\) and \(n\).
Solution
For a field \(F\), the singular chain complex \(C_*(X;F)\) is a chain complex of vector spaces, so dualization is exact. Hence \[ H^n(X;F)\cong \operatorname{Hom}_F(H_n(X;F),F). \] Thus the hypothesis implies \[ \widetilde H_n(X;\mathbb Q)=0, \qquad \widetilde H_n(X;\mathbb Z_p)=0 \] for every \(n\) and every prime \(p\).
By the homology universal coefficient theorem, \[ 0\to \widetilde H_n(X;\mathbb Z)\otimes\mathbb Z_p \to \widetilde H_n(X;\mathbb Z_p) \to \operatorname{Tor}(\widetilde H_{n-1}(X;\mathbb Z),\mathbb Z_p) \to0. \] Since the middle group is zero for every prime \(p\), one has for \[ A_n=\widetilde H_n(X;\mathbb Z) \] both \(A_n/pA_n=0\) and \(A_n[p]=0\) for every prime \(p\). Equivalently, multiplication by every prime is bijective on \(A_n\), hence multiplication by every nonzero integer is bijective. Therefore each \(A_n\) is naturally a vector space over \(\mathbb Q\).
But \[ 0=\widetilde H_n(X;\mathbb Q)\cong A_n\otimes\mathbb Q. \] For a \(\mathbb Q\)-vector space \(A_n\), the canonical map \(A_n\to A_n\otimes\mathbb Q\) is an isomorphism, so \(A_n=0\). Thus \[ \widetilde H_n(X;\mathbb Z)=0 \] for all \(n\).
Finally the universal coefficient theorem for cohomology gives \[ 0\to \operatorname{Ext}(\widetilde H_{n-1}(X;\mathbb Z),G) \to \widetilde H^n(X;G) \to \operatorname{Hom}(\widetilde H_n(X;\mathbb Z),G)\to0, \] and both outer groups vanish. Hence \[ \widetilde H^n(X;G)=0 \] for every coefficient group \(G\) and every \(n\).