Let $G$ be a group and $H$ , $K$ be subgroups of $G$ such that $[G:H]$ and $[G:K]$ are finite. Then is it true that $[G:H∩K]$ is also finite ?

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    $\begingroup$ The answer is yes, as is proved in many different ways here. Anon's second proof is fairly slick. $\endgroup$ – Jared Apr 24 '13 at 5:44
  • $\begingroup$ @Jared: Thanks. $\endgroup$ – Souvik Dey Apr 24 '13 at 5:50

If $(H\cap K)a$ be a right coset of $H\cap K$ in the group, try to show that $(H\cap K)a=Ha\cap Ka$. What does it mean? It means that every right coset of $H\cap K$ is an intersection of an right coset of $H$ with the right coset of $K$. In fact; every right coset of $H\cap K$ has a form $Hx\cap Ky$ and so whole number of such these right cosets which are distinct is equal or less than the number of such combinations above. That means that $$[G:H\cap K]\leq[G:K][G:H]$$ The problem gets beautiful when the indices of two subgroup in $G$ are coprime.

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    $\begingroup$ Yes, indeed, when the indices of two subgroups are coprime, it does get beautiful! Nice guidance! $\endgroup$ – amWhy Apr 25 '13 at 1:56

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