Mutliplication table of quotient groups

gotmilk04
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Homework Statement


Write the multiplication table of C_{6}/C_{3}
and identify it as a familiar group.


Homework Equations





The Attempt at a Solution


C_{6}={1,\omega,\omega^2,\omega^3,\omega^4,\omega^5}
C3={1,\omega,\omega^2}
The cosets are C3 and \omega^3C3
I just need help making the multiplication table.
 
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I'm assuming C_n and C^n both refer to the cyclic group of order n, since that's the impression I get from your post.

if you meant for C_6 to be generated by \omega, then you should have C_3 = \{1,\omega^2,\omega^4\} because otherwise C_3 is not a group. Then the cosets should be C_3, \omega C_3.

What exactly are you having trouble with? As you said yourself the group C_6/C_3 has exactly two elements (C_3 and \omega C_3), so the following four are the possible products you need to compute and insert in the multiplication table:
C_3 \times C_3
C_3 \times \omega C_3
\omega C_3 \times C_3
\omega C_3 \times \omega C_3
 
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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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