Prove that any finite abelian group that can act freely on $S^n$ is cyclic.
My attempt
The fundamental theorem of finite abelian groups states that every finite abelian group $G$ can be expressed as the direct sum of cyclic subgroups of prime-power order. Then it's not hard to show that any finite abelian group $G$ which isn't cyclic has a subgroup $\mathbb{Z}_p\oplus\mathbb{Z}_q$, where $p$ and $q$ are distinct prime numbers.
Since if $G$ can act freely on $X$, any subgroup of $G$ can act freely on $X$. It suffices to prove that $\mathbb{Z}_p\oplus\mathbb{Z}_q$ can't act freely on $S^n$.
Write $G=\mathbb{Z}_p\oplus\mathbb{Z}_q$. In my previous post I proved this proposition for $n=1$ and $n=2k$. For $n\ge3$, an application of Lefschetz shows that $M:=S^n/G$ is a closed orientable manifold. Furthermore for $n\ge3$ we know that $H_1(M)\cong \pi_1(M)\cong G$ and $H_1(M;\mathbb{Z}_p)\cong H_{n-1}(M;\mathbb{Z}_p)$.
Now, $\pi_k(M) = 0$ for $k = 2,..., n-1$. Hence we can turn $M$ into an Eilenburg-Maclane space $K(G,1)$ by attaching cells of dimension $n+1$ or higher, which has the same homotopy type of $L_p\times L_q$ where $L_p$ and $L_q$ are two infinite dimensional lens spaces. Then we have $H_{n-1}(L_p\times L_q;\mathbb{Z}_p)\cong H_{n-1}(M;\mathbb{Z}_p)\cong H_{1}(M;\mathbb{Z}_p) \cong G \otimes \mathbb{Z}_p \cong \mathbb{Z}_p$.
But from Kunneth theorem we have $H_{n-1}(L_p\times L_q;\mathbb{Z}_p)\cong \bigoplus_{i+j=n-1} H_{i}(L_p;\mathbb{Z}_p) \otimes H_{j}(L_q;\mathbb{Z}_p)\tag{*}$
Note that $H_k(L_m) = \begin{cases} \mathbb{Z}, & \mbox{if }k=0\\ \mathbb{Z}_m, & \mbox{if }k>0\mbox{ is odd} \\ 0, & \mbox{if }k>0\mbox{ is even} \end{cases}$. Thus $(*)$ lead to a contradiction.
Is this right? And could you give me some hints about whether we can prove the proposition for infinite ablelian groups? Thanks in advance!