This question/thread is continuation from here Also I.N Herstein in Page 40 Question 6 has brought up the same question
My approach: We know the property $\left | G:K \right |\leq \left | G:H \right|\left | H:K \right | $ if $K$ is the subgroup of $H$ which is the subgroup of $G$
So if we can prove $H\cap K$ is a subgroup of H, then the work is done
We know, $H \cap K \subseteq H$
$\Rightarrow$ if $h_1,h_2 \in H\cap K$, then $h_1,h_2 \in H$ also if $h_1,h_2 \in H\cap K$, then $h_1,h_2 \in K$
$\Rightarrow$ $h_1\cdot h_2 \in H$ and $h_1\cdot h_2 \in K$
$\Rightarrow$ $h_1\cdot h_2 \in H \cap K$
Thus its closed under product
Similarly, we can argue about $1_G \in H \cap K $ [since $H,K$ are both subgroups and thus identity must be common to both of them]
$\Rightarrow$ inverse exists in $H \cap K$
Thus $H \cap K$ is a subgroup of $H$ as well as $K$
Hence applying the property
$\left | G:K \right | \leq \left | G:H \right|\left | H:K \right | $
We get if $\left | G:H \right|$ = $m$ [since finite] and $\left | H:H \cap K \right|$=$k$
$\Rightarrow \left | G:K \right | \leq mk$
Equality holds when $G$ is a finite group, $mk$ being the upper bound of the index of $H\cap K$ in $G$
Soham
Please do raise any questions if I have gone wrong somewhere
Soham