This question is from the counting principle section of my abstract algebra textbook. I understand the proof that is provided in my textbook. The proof uses a counterexample with subgroups $H$ and $K$ to get a contradiction with $o(H\cap K)=o(H)=o(K)$. I was wondering if this was an alternative proof.
Suppose there are two subgroups with order $p$. We know that $$o(H\cap K)\ge\frac{o(H)o(K)}{o(G)}>1$$
The middle equation would be $\frac{p^2}{pq}=\frac{p}{q}$. Since $p$ is a prime and $p>q$, $\frac{p}{q}$ wouldn't be an integer.
If I am wrong, I think I am incorrectly assuming that $\frac{p}{q}$ has to be an integer, but it feels like it has to, intuitively.
Let $H$, $K$ be subgroups of order $p$. Observe that the intersection $H\cap K$ of two subgroups is again a subgroup, and therefore its order divides $p$. But $p$ is prime, so we are left with two possibilities: either $o(H\cap K) = p$, in which case $H = K$ and we are done. The other case is $o(H\cap K) = 1$, which means the intersection of $H$ and $K$ is trivial. But then we have $o(HK) = p² > pq = o(G)$, a contradiction. (Note that $HK$ is seen only as a subset of $G$, not a subgroup).
Your proof leads to the same conclusion: If $o(H\cap K) > 1$, then it must be equal to $p$, and therefore $H = K$. It does not matter if the fraction evaluates to an integer, but I would like to know where the formula comes from.