Upper bound for index $|G:H\cap K|$

This question is already answered here prove that $H\cap K$ have finite index in G
But I had written solution in another way.
Let $G$ be group and $H$ and $K$ be its subgroup with finite index then we have to show bounds for index of $H \cap K$ as $[a,b]\leq |G:H\cap K|\leq ab$ where $a=|G:H|$ and $b=|G:K|$
$|G:H\cap K|=|G:H||H:H\cap K|=|G:K||K:H\cap K|$
As $a$ and $b$ both divide $|G:H\cap K|$, we have $[a,b]$ divide $|G:H\cap K|$ that is $[a,b]\leq |G:H\cap K|$
Now the only thing that remains is to show that $|G:H\cap K|\leq ab$
This can be simplied to show by $|G:H\cap K|=|G:H||H:H\cap K|\leq |G:H||G:K|$
That is $|H:H\cap K|\leq |G:K|$
That is coset of K in G are more than Coset of $H \cap K$ in H .But How to Prove that? How to proceed further? Any Help will be appreciated

• You seem to have proved it. What is your question? – Derek Holt Jul 16 '18 at 11:00
• I am not able to write answer of last claim second last line – MathLover Jul 19 '18 at 6:54
• @DerekHolt . coset of K in G are more than Coset of $H \cap K in H .But How to Prove that?Sir Please Help me I am still Stuck – MathLover Jul 28 '18 at 13:15 • If$h_1,h_2 \in H$with$(H \cap K)h_1 \ne (H \cap K)h_2$, then$Kh_1 \ne Kh_2$, so$|H:H \cap K| \le |G:K|$. – Derek Holt Jul 28 '18 at 14:36 • Thanks a Lot Sir Now I understand – MathLover Jul 28 '18 at 16:36 1 Answer Here is one way to do it. Let$x_1H, \ldots, x_aH$be all the distinct left cosets of$H$in$G$, and$y_1K, \ldots, y_bK$be all the distinct left cosets of$K$in$G$. To show that$|G: H\cap K|\leq ab$, it suffices to show that whenever we have$z_1, \ldots, z_{ab+1}\in G$, then$z_iH\cap K=z_jH\cap K$for some$i\neq j$. To see this, we have by the pigeonhole principle that some$z_{i_1}, \ldots, z_{i_{b+1}}$are in the same left coset of$H$. Renumbering these for convenience, we may say that$z_1, \ldots, z_{b+1}$are in the same left coset of$H$. Again, by the pigeonhole principle, some two of$z_1, \ldots, z_{b+1}$, say$z_i$and$z_j$with$i\neq j$, are in the same left coset of$K$. So we have$z_i$and$z_j$are in the same left coset of$H$and the same left coset of$K$. Thus$z_j^{-1}z_i\in H\cap K$, and consequently we have$z_iH\cap K=z_j H\cap K\$.

This finishes the proof.