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Similar question has been asked on SE before but the problem statement is usually more specific and gives more information (in particular, tells you what to prove), but this problem asks to prove or disprove: For a group $G$ and subgroups $H$ and $K$, if $|G:H|$ and $|G:K|$ are finite then so is $|G:H \cap K|$.

Is there an easy way to see this is true with given just statement?

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    $\begingroup$ $H\cup K$? What about the title? (Did they change the formatting on the \cup ? It looks more squarish.) $\endgroup$
    – Eoin
    Jun 27, 2015 at 23:41
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    $\begingroup$ You beat me to my edit by about 12 seconds. I did cup instead of cap on accident. $\endgroup$
    – Tuo
    Jun 27, 2015 at 23:43
  • $\begingroup$ Here's a hint: Any coset of $H\cap K$ is the intersection of a coset of $H$ and a coset of $K$. $\endgroup$ Jun 27, 2015 at 23:48
  • $\begingroup$ Note that $[H:H\cap K] \leq [G:K]$ and $[G:H\cap K]=[G:H][H:H\cap K]$. $\endgroup$
    – Taylor
    Jun 28, 2015 at 0:03

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Okay I think I got it. As Meelo stated, we have $g(H \cap K)=gH \cap gK$. To see this if $x \in g(H \cap K)$ then $x=gs$ for some $s \in H \cap K$. But then $x=gs \in gH$ and $x=gs \in gK$ so $x \in gH \in gK$ so $g(H \cap K) \subseteq gH \cap gK$. Similarly if $x \in gH \cap gK$ then $x=gs$ for $s \in H$ and $x=gt$ for $t \in H$. But then $s=t$ so $x \in g(H \cap K$ so $g(H \cap K) \supseteq gH \cap gK$. Hence any coset of $H \cap K$ is is the intersection of a coset of $H$ and and $K$ so we have $|G:H \cap K| \leq |G:H||G:K|$ which is finite by assumption.

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