# normal subgroup of a non abelian group with index $n$. [closed]

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Krish, T. Bongers, Derek Holt, Maria Mazur
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• $n=-1$?${}{}{}{}$ – Dave Nov 22 '17 at 4:30
• 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. – Dave Nov 22 '17 at 4:31
• Maybe you could help with some hint :) – 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 $G$?