Index of intersection of two subgroups

Question

Let $H$ and $K$ be subgroups of $G$ with indices $3$ and $5$ in $G$.

I need to show that the index of $H\cap K$ is a multiple of $15$.

ATQ

$\frac{|G|}{|H|}$ = 3

So $|G|$ is a multiple of $3$.

Similarly, $|G|$ is a multiple of $5$.

So we conclude that $|G|$ is a multiple of $15$.

Now $H\cap K$ is a subgroup of $G$. So by Lagrange's theorem , it will divide $|G|$.

From here, how can I conclude that its order is a multiple of $15$?

Answer 1

Hint. $[A : C] = [A : B] [B : C]$ for (finite) groups $A$ with subgroups $B ⊆ A$ and $C ⊆ B$.