Group Theory - specific non-abelian case

audiowize
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Homework Statement



Let G be an Abelian group and let H+{x^3 : x is an element of G}

Find a non-Abelian group in which H is not a subgroup

Homework Equations



I wish it was that easy...

The Attempt at a Solution



I looked at the quaternion group, and some other matrix groups, but no luck so far...
 
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What is the first non-abelian group in any reasonable ordering?
 
The permutation group S_3.

I looked at this also... but the cubes of the individual elements seemed closed under the operation of composition, and the identity is present... Four of us have spent about 3 hours on this without much luck, any nudging is greatly appreciated!
 
You should triple-check your work, because matt is never wrong. If G=S_3, what are you thinking H is?
 
Thanks for the help guys! I'm sure glad I don't have to take exams at midnight. A fresh set of eyes in the morning and your confirmation sure helped!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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