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I have this problem:

For a group $ G $ and a subgroup $H$ which is not normal, show there exists a prime $p$ such that $p \mid |G| $ and $ p < [G:H] $

My best approach is to maybe show that $p$ is dividing $[G:H] $, and that $ p = [G:H] $ is not possible due to $H$ being not normal. However, I might be missing some technical stuff or theorems regarding groups/number theory and I don't really know how to progress here.

Keep in mind that I still don't really grasp group theory, so it might be trivial and I'm missing something. Anyway, I would like some hint or approach to this.

Thanks for the help!

Shaun
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CauchyChaos
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    Perhaps this is easier to think about if you rephrase it as "if a subgroup has index $p$, the least prime dividing the order of the group, then the subgroup is normal." – lulu Nov 12 '21 at 14:46
  • Use $a\mid b$ for $a\mid b$ – Shaun Nov 12 '21 at 14:50
  • There's something I don't understand. If what you're saying is true, then if $H$ is not normal, and $p$ is the smallest prime such that $p \mid |G|$, then it means that $[G:H]$ is not equal to $p$. But can it be that $[G:H] < p $? – CauchyChaos Nov 12 '21 at 14:57
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    @LagrangeLover: Any prime that divides $[G:H]$ would be smaller than $p$ and necessarily divide $|G|$ (since $[G:H]$ divides $|G|$), but $p$ is supposed to be the smallest prime that divides $|G|$. – Arturo Magidin Nov 12 '21 at 15:01

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