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The title speaks for itself, I saw something that requires it but I can't seem to prove it, any help is appreciated!

I'm asking because of this comment which specify that from the second isomorphism theorem we know that $[HK:H]=[K:H\cap K]$ suggesting that $H\triangleleft HK$ when in the question we have that $K\triangleleft G$ and $H\le G$

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Kim Seel
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    Let $G$ be a group with a nonnormal subgroup $H$. Let $N=G$. Then $HN=G$ and $H$ is not normal in $HN$. Are you sure that this is what you want? – martin.koeberl Mar 23 '17 at 14:51
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    If you pick $N=G$ then it would mean that every subgroup is normal. Which is not true. So, there are plenty of counterexamples. Even with $N$ a proper subgroup of $G$. Try some small examples. – Nicky Hekster Mar 23 '17 at 14:52
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    I'm asking because of this – Kim Seel Mar 23 '17 at 14:52
  • What do you don't understand about the proof? Make that clear to us please, so we can help. – Nicky Hekster Mar 23 '17 at 14:55
  • Edit the link into your question. It is absurd to omit such information when you have it. What part exactly do you think requires this (false) result? – rschwieb Mar 23 '17 at 14:55
  • I think I clarified by your comments I now know that it's false just as I suspected but it's still not clear to me how it "adds up" – Kim Seel Mar 23 '17 at 15:05

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It appears to me that your core mistake is that you seem to think the 'index of' relation (denoted by the '[:]') is only applicable between a group and a normal subgroup. Actually it is applicable between a group and any subgroup.

PMar
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