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Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $H\cap K$ also a subgroup of finite index? I feel like need that $[G\colon(H\cap K)]$ is a divisor of $[G\colon H]\cdot[G\colon K]$, but I dont't know when this holds.

Can somebody help me out?

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    Step 1: Show that $H \cap K$ is a subgroup of $H$ of finite index. Step 2: Show that $[G:H] = [G:H \cap K][H\cap K : H]$.2012-04-05
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    @MartQ. You don't have to show that it's a divisor. It suffices to show that $[G\colon(H\cap K)]\leq [G\colon H]\cdot[G\colon K].$ You can try to find an injection from a certain set to another.2012-04-05
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    [A related question.](http://math.stackexchange.com/questions/86086/leftgh-cap-k-right-leftgh-right-leftgk-right-if-leftgh-right)2012-04-05
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    @CliveNewstead The formula $[G:H] = [G:H \cap K][H\cap K : H]$ seems incorrect to me. What is $[H\cap K : H]?$ Did you mean $[G:H\cap K]=[G:H][H:H\cap K]?$2012-04-05
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    @ymar: Yes, sorry, I was having a dense moment. Unfortunately it's too late to edit my comment.2012-04-07
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    Proof of Eklavya in https://math.stackexchange.com/questions/1605121/prove-that-h-cap-k-have-finite-index-in-g: Awra Dip uses $g(H\cap K) = gH \cap gK$, but proof of Eklavya uses only $g(H\cap K) \subseteq gH \cap gK$. Since there are only finitely many choices for $gH$ and for $gK$, there are only finitely many choices for $g(H \cap K)$.2018-10-14

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