prove that $H\cap K$ have finite index in G

If $G$ is a group and $H,K$ are two subgroups of finite index in $G$, prove that $H\cap K$ is of finite index in $G$. Can you find upper bound for index of $H\cap K$ in $G$?

Since $a(H\cap K$) $\subseteq aH\cap aK$, and since there are only a finite many choices for $aH$ and $aK$, there are only finite many choices for $a(H\cap K$)

Is it correct ?

What about upper bound of index of $H\cap K$?

If you are comfortable with group actions: Let G act on the set $S=G/H \times G/K$ (as coset spaces) by the law $g \star (aH,bK)=(gaH,gbK)$.Consider the element $(H,K)$ of $S$. Stab($(H,K)=H \cap K)$. Now we know that $[G: Stab((H,K)]=|Orb((H,K))| \leq |S|=[G:H][G:K]$. This upper bound is achievable (not too hard to find an example).

By multiplicativity of the subgroup index, we have $$ [G: H \cap K] = [G : K] [K : H \cap K] \, . $$ We claim that $[K : H \cap K] \leq [G : H]$. Define a map \begin{align*} f: \frac{K}{H \cap K} &\to \frac{G}{H}\\ k (H \cap K) &\mapsto kH \, . \end{align*} It is straightforward to show that $f$ is well-defined. We now show $f$ is injective. Suppose $k_1 H = f(k_1(H \cap K)) = f(k_2(H \cap K)) = k_2 H$. Then $k_2^{-1} k_1 \in H$, so $k_2^{-1} k_1 \in H \cap K$, hence $k_1(H \cap K) = k_2(H \cap K)$ as desired. Thus $f$ is injective, so $[K : H \cap K] \leq [G : H]$. Therefore $$ [G: H \cap K] = [G : K] [K : H \cap K] \leq [G : K] [G : H] \, . $$ Since the indices on the righthand side are finite by assumption, this shows that the index of $H \cap K$ is also finite.