Let $n$ be an postive integer. Prove that there exists is a non abelian finite group containing a normal subgroup of index $n$.

Please, some hint.


closed as off-topic by Leucippus, Krish, T. Bongers, Derek Holt, Maria Mazur Nov 22 '17 at 8:45

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  • $\begingroup$ $n=-1$?${}{}{}{}$ $\endgroup$ – Dave Nov 22 '17 at 4:30
  • $\begingroup$ But seriously though: you should include your attempts at solving the problem so that the people on this site can help you with specific problems you have. Most people don't want to just do other people's homework. $\endgroup$ – Dave Nov 22 '17 at 4:31
  • $\begingroup$ Maybe you could help with some hint :) $\endgroup$ – Javiera G Nov 22 '17 at 5:08

Hint: Take your favorite non-abelian group $G$, and take some other group $H$. Then the product group $G\times H$ is non abelian, and $G$ sits inside as a normal subgroup $G\times \{1\}$. What is the index of $G$?

  • $\begingroup$ is the orden of H. Thanks! $\endgroup$ – Javiera G Nov 22 '17 at 5:22

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