From the l.e.s. of homology groups associated to the s.e.s. of chain complexes $0\to C_i(X)\xrightarrow{n}C_i(X)\to C_i(X;\Bbb Z_n)\to 0$ deduce immediately that there are short exact sequences
$$0\to H_i(X)/nH_i(X)\to H_i(X;\Bbb Z_n)\to n\text{-Torsion}(H_{i-1}(X))\to 0$$
where $n\text{-Torsion}(G)$ is the kernel of the map $G\xrightarrow{n}G,g\mapsto ng$. Use this to show that $\tilde{H}_i(X;\Bbb Z_p) = 0$ for all $i$ and all primes $p$ iff $\tilde{H}_i(X)$ is a vector space over $\Bbb Q$ for all $i$.
I understood we can get a s.e.s. But some minor problem for reduced homology case. If $i>1$ then no problem. If $i=1$ then still ok because $H_0(X)$ if free group. If $i=0$, then still true? (of course it should be true but why?).
Best Answer
You should just use the short exact sequence for non-reduced homology, and that's all the exercise asks you to establish. For the second part, the only way for $\widetilde{𝐻}_0(𝑋)$ to be a vector space over $\mathbb{Q}$ is if $X$ is path-connected; if not, $\widetilde{𝐻}_0(𝑋)$ is a nontrivial free abelian group, and that can't be a rational vector space.