Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $H\cap K$ also a subgroup of finite index? I feel like need that $[G\colon(H\cap K)]$ is a divisor of $[G\colon H]\cdot[G\colon K]$, but I dont't know when this holds.
Can somebody help me out?
Best Answer
Proof $1$: $\quad[G:H\cap K]=[G:H][H:H\cap K]=[G:H][HK:K]\le [G:H][G:K].$
We do not assume normality on $H,K$ and therefore cannot assume $HK$ is a group. But it is a disjoint union of cosets of $K$, so the index makes sense. Also, $[H:H\cap K]=[HK:K]$ follows from the orbit-stabilizer theorem: $H$ acts transitively on $HK/K$ and the element $K$ has stabilizer $H\cap K$.
Proof $2$: Consider the diagonal action of $G$ on the product of coset spaces $G/H\times G/K$. The latter is finite so the orbit of $H\times K$ is finite, and the stabilizer of it is simply $H\cap K$. Invoke orbit-stabilizer.