I've been considering this problem: Suppose that $X$ is a topological space and that $H_1(X)$ is a finite group of odd order. Show that if $p:\tilde{Y}\rightarrow Y$ is a covering space of index $2$, then any map $f:X\rightarrow Y$ can be lifted to a map $\tilde{f}:X\rightarrow \tilde{Y}$ so that $p\circ \tilde{f}=f$.
It is easy to show that any index $2$ subgroup contains all elements of odd order. So, if $\gamma\in\pi_1(X)$ has odd order, then $f_*(\gamma)\in p_*(\pi_1(\tilde{Y}))$. So, I'm thinking that it might be if $H_1(X)$ is finite of odd order, $\pi_1(X)$ must have all elements of odd order. Is that so? Or, is there another way to do this problem?