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How to prove: Let $H_1,\cdots,H_n$ be subgroups with finite indices of group $G$. Then $G$ has a normal subgroup $N$ of finite index such that $N\leq H_i$ for each $i$.

I do not know how to prove it. Even for the case $n=1$, after letting $N=\bigcap_{g\in G}gH_1g^{-1}$, I am confused on how to prove the index of $N$ to be finite.

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  • $\begingroup$ Are u familiar with "an action"? I can do the problem while $n=1$ by using actions. $\endgroup$ – mrs Sep 3 '13 at 6:25
  • $\begingroup$ It is also worth observing that the intersection of two subgroups of finite index has finite index, so you can reduce to the case $n=1$. $\endgroup$ – Derek Holt Sep 3 '13 at 8:10
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Consider the action of $G$ on the set $X=(G/H_1)\times\dots\times (G/H_n)$. The set is finite. Let $N$ be the kernel of this action (i.e. of the morphism $G\to S_X$).

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