isomorphic homotopy groups, but not homotopy equivalent

Question

We are supposed to show that the spaces $X= \mathbb{R}P^2$ and $ Y= S^2 \times \mathbb{R}P^\infty$ have isomorphic homotopy groups but are not homotopy equivalent.
I already showed that all homotopy groups are isomorphic, but struggled to find a reason why $$X \sim_h Y $$ cannot be the case. Any help is appreciated. We have not introduced homology so far.

Answer 1

They can be distingushed by the action of $\pi_1 \cong \mathbb{Z}_2$ on $\pi_2 \cong \mathbb{Z}$. This action is trivial for $S^2 \times \mathbb{RP}^{\infty}$ but is given by inversion for $\mathbb{RP}^2$, because the antipode map on $S^2$ acts by $-1$ on $\pi_2$.