Suppose you have a group $G$ of order $ms$ and it is $m \leq p$ for every prime factor $p$ of $s$. We want to show that a subgroup $H$ with $(G:H)=m$ is a normal subgroup of $G$.
So let $H$ be a subgroup of $G$ with $\vert H\vert=s$ and $K$ be another subgroup of $G$ of order $s$. Then we have
$$ \vert HK\vert= \frac{\vert H \vert \vert K \vert}{\vert H \cap K \vert}=\frac{s^2}{\vert H \cap K \vert} \leq ms \Longleftrightarrow \frac{s}{\vert H \cap K\vert} \leq m.$$
Suppose now that $\frac{s}{\vert H \cap K\vert}>1$. Is it always true that in this case $m$ is the smallest prime divisor of the group order? I read this in another post, but I don't see an argument for this.
My attempt:
$\vert H \cap K \vert$ is a divisor of $s$ (how to show?). Then $m$ is the smallest prime divisor of $\frac{s^2}{\vert H \cap K\vert}$. But how do we get to the order of $G$ in this case?
Maybe there is an easier way to prove this statement...
You were able to prove that $$\frac{s}{\vert H \cap K\vert} \leq m.$$
Note that by Lagrange's Theorem and since $H\cap K$ is a subgroup of $H$, $\frac{s}{\vert H \cap K\vert}$ is actually an integer (which was one of your questions). Therefore if you write down the prime factorization of $s$, $\vert H \cap K\vert$ is a product of some of these primes and $\frac{s}{\vert H \cap K\vert}$ is the complementary product of the remaining primes:
If $s=p_1^{\alpha_1}\ldots p_k^{\alpha_k}$ then $\frac{s}{\vert H \cap K\vert} =p_1^{\beta_1}\ldots p_k^{\beta_k}$ for some $0\leq\beta_i\leq \alpha_i$.
So you have $$\underbrace{p_1^{\beta_1}\ldots p_k^{\beta_k}}_{\text{Thereafter called LHS}}\leq m\leq \min_{i=1,\ldots , k}\{p_i\}$$ where the second inequality is from the fact that [being less than or equal to every element in a finite list] is equivalent to [being less than or equal to the minimum element from that list].
From there, it is clear that the LHS can either be equal to $1$ or to $\min_{i=1,\ldots , k}\{p_i\}$, otherwise it would violate transitivity of $\leq$.
If you assume $\frac{s}{\vert H \cap K\vert}>1$ then the LHS is not $1$ and it must be $m$. Therefore $m=\min_{i=1,\ldots , k}\{p_i\}$.