I am currently studying for an algebra qualifying exam and am stuck on how to even start this problem.
Suppose that a group $G$ acts transitively on sets $X$ and $Y$, where $1 < |X| < |Y| = p$ and $p$ is prime. Show that $G$ is not simple.
I found this problem: Let group $G$ acts transitively on finite sets $\Omega$ and $\Psi$, with $1<|\Omega|<|\Psi|$. If $G$ is simple, show that $|\Psi|$ cannot be prime., which asks you to show that if $G$ is simple, then $|Y|$ is not prime. I've tried reading through it to see if I could gain some insight to help with this problem, but I'm still stuck.
I tried a proof by contradiction (assume $G$ is simple) and found that the only normal subgroup of $G$ must be the trivial group, or else if it were $G$, then $G$ wouldn't act transitively on $X$ and $Y$. Using the page I linked, you can also show that $G$ acts faithfully on $X$ and $Y$. Then I got stuck and have no idea how to proceed.
Another approach is that we know $p = \frac{|G|}{|G_Y|}$ and $|X| = \frac{|G|}{|G_X|} < p$. The page I linked stated that "if $p$ is prime then clearly $p$ cannot be the smallest prime dividing $|G|$, i.e., we must have another prime $q<p$ with $q \mid |G|$. I don't understand this reasoning, but I think it may help with finding a non-trivial normal subgroup, since if $q \mid |G|$, then we have a subgroup of order $q$.
Any suggestions would be great!