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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

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  • $\begingroup$ You seem to have proved it. What is your question? $\endgroup$ – Derek Holt Jul 16 '18 at 11:00
  • $\begingroup$ I am not able to write answer of last claim second last line $\endgroup$ – MathLover Jul 19 '18 at 6:54
  • $\begingroup$ @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 $\endgroup$ – MathLover Jul 28 '18 at 13:15
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    $\begingroup$ 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 $|H:H \cap K| \le |G:K|$. $\endgroup$ – Derek Holt Jul 28 '18 at 14:36
  • $\begingroup$ Thanks a Lot Sir Now I understand $\endgroup$ – MathLover Jul 28 '18 at 16:36
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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.

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