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Let $H$ and $K$ be finite index subgroups of a group $G$ with index $h$ and $k$, respectively.

I know that $H\cap K$ is of finite index in $H$ and $K$.

Is the index of $H\cap K$ in $H$ bounded by $k$?

By symmetry the index of $H\cap K$ in $K$ would be bounded by $h$.

In standard notation: does the inequality

$$[H:H\cap K] \leq [G:H]$$ hold? Edit: I meant to write $[H:H\cap K] \leq [G:K]$.

There are many related MS questions, but I don't see how they answer this question.

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  • $\begingroup$ The suggested inequality $[H:H\cap K]\le [G:\color{red}{H}]$ is certainly not justified. $\endgroup$ – Hagen von Eitzen Oct 23 '14 at 10:34
  • $\begingroup$ Consider $G=\Bbb{Z}$, $H=2\Bbb{Z}$ and $K=6\Bbb{Z}$. $\endgroup$ – Hanul Jeon Oct 23 '14 at 10:37
  • $\begingroup$ Thank you for your comments! $\endgroup$ – Stephane Oct 23 '14 at 11:14
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The map $$ \begin{align}H/(H\cap K)&\to G/K\\ g(H\cap K)&\mapsto gK\end{align}$$ is injective because $gK=hK$ with $g,h\in H$ implies $h^{-1}g\in K$ and hence also $\in H\cap K$.

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  • $\begingroup$ Thank you! This answers my (corrected) question in the affirmative, as I indeed meant to write $[G:K]$. $\endgroup$ – Stephane Oct 23 '14 at 11:16

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