Questions about Hatcher 3.2.16 Hatcher's exercise 3.2.16 states

Show that if X and Y are finite CW complexes such that $H^*(X;\mathbb{Z})$ and $H^*(Y;\mathbb{Z})$ contain no elements of order a power of a given prime p, then the same is true for $X \times Y$.

Now the hint says to use the Künneth Formula

$H^*(X;R)\otimes_R H^*(Y;R) \cong H^*(X \times Y ;R)$ when X,Y are CW complexes and $H^k(Y;R)$ is a finitely generated free R module.

for various fields R.
Here's what I need help with (my attempts are also included):

*

*Firstly, I'm looking for a clarification about the question. I know the following logic is flawed but I can't figure out why:
Since X is a finite CW complex, $H^k(X,R) = 0$ for all $k \geq m$ for some number $m$ since there are only finitely many cells. So for $\alpha$ in $H^1(X)$, we have that $\alpha^m \in H^m(X;R) = 0$ (product given by the cup product $\alpha \smile \alpha)$. So pick $n$ such that $p^n > m$, then $\alpha^{p^n} = 0$ and $H^*(X)$ has an element of order $p^n$ always. I'm aware that I'm probably missing something basic.


*Assuming the question is well posed and that 1) is not an issue, then my first thought was to use the formula not for a field but with $R = \mathbb{Z}$. If $H^*(X;\mathbb{Z})$ and $H^*(Y;\mathbb{Z})$ have no elements of order $p^n$ for any n, then my first guess is that they are torsion free. If they are indeed torsion free, then by the Universal Coefficient Theorem, we have that $H^*(X;\mathbb{Z}) \cong \mathbb{Z}^r$ and $H^*(Y;\mathbb{Z}) \cong \mathbb{Z}^l$ so $H^*(X\times Y;\mathbb{Z}) \cong H^*(X;\mathbb{Z}) \otimes H^*(Y;\mathbb{Z}) \cong \mathbb{Z}^{rl}$ which is torsion free. But I haven't been able to show why $H^*(X;\mathbb{Z})$ would need to be torsion free. If this is correct, then I'm looking to see why that is the case, and if not, then a counterexample.


*Assuming I was wrong and the approach  in 2) was incorrect, I tried thinking about Hatcher's hint more carefully and then decided to look at $H^*(X;\mathbb{Z}/p\mathbb{Z})$ as a $\mathbb{Z}/p\mathbb{Z}$ module. I couldn't get very far but the hope was to extend $H^*(X;\mathbb{Z}) \rightarrow H^*(X;\mathbb{Z}/p\mathbb{Z})$. My guess being that the latter is a free $\mathbb{Z}/p\mathbb{Z}$ module and by the Künneth formula, $H^*(X \times Y;\mathbb{Z}/p\mathbb{Z}) \cong H^*(X;\mathbb{Z}/p\mathbb{Z}) \otimes H^*(Y;\mathbb{Z}/p\mathbb{Z})$, which is an extension of $H^*(X \times Y;\mathbb{Z})$ and was hoping to construct an order $p^n$ element here towards a contradiction. No luck so far, but any advice would be great.


*I also found this MSE post which I didn't find too helpful. The comments seem to suggest studying the order of elements or use Lagranges Theorem in the tensor product but wouldn't we expect the tensor to be free over $\mathbb{Z}$ and have elements of infinite order and Lagranges theorem only applies to finite groups?
 A: If $X$ is a finite CW complex, then its rational homology $H^n(X; \mathbb{Q})$ is a rational vector space. By the Universal Coefficient Theorem (Theorem 3.2 in Hatcher, applied with $G=\mathbb{Q}$), its dimension is the rank of $H_n(X)$ as a free abelian group: the finite summands of $H_n(X)$ do not contribute (since there are no maps from a finite group into $\mathbb{Q}$), and nothing from $H_{n-1}(X)$ contributes (since $\mathbb{Q}$ is injective as an abelian group, or you can use the properties of Ext that Hatcher gives just after the theorem to see that the Ext group is zero when $G=\mathbb{Q}$).
If $p$ is a prime, the only way for $H^n(X)$ to have elements of order a power of $p$ is for $H_n(X)$ to also have elements of order a power of $p$, this time by the Universal Coefficient Theorem with $G=\mathbb{Z}$. So if $H^n(X)$ has no elements of order $p^m$ for any $m$, then the same is true for $H_n(X)$. In this case, $H^n(X; \mathbb{F}_p)$ is an $\mathbb{F}_p$-vector space with dimension again equal to the rank of $H_n(X)$ as a free abelian group: any summands of order prime to $p$ in $H_n(X)$ will not contribute to $H^n(X; \mathbb{F}_p)$ by the Universal Coefficient Theorem (with $G=\mathbb{Z}/p$).
Now by the above and the Künneth theorem with various coefficients, the dimension of $H^n(X \times Y; \mathbb{F}_p)$ is equal to
\begin{align}
\dim H^n(X \times Y; \mathbb{F}_p) &= 
\sum_{i=0}^n \dim H^i(X; \mathbb{F}_p) \dim H^{n-i}(Y; \mathbb{F}_p) \\
&= \sum_{i=0}^n rk H_i(X) rk H_{n-i}(Y) \\
&= \sum_{i=0}^n \dim H^i(X;\mathbb{Q}) \dim H^{n-i}(Y; \mathbb{Q}) \\
&= \dim H^n(X \times Y; \mathbb{Q}) \\
&= rk H_n(X \times Y).
\end{align}
On the other hand, if $H^*(X \times Y)$ contained an element of order $p^m$ for some $m$, then the same is true for $H_*(X \times Y)$, and so the dimension of $H^n(X \times Y; \mathbb{F}_p)$ will not equal the rank of $H_n(X \times Y)$ for some $n$.
