Mutliplication table of quotient groups

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SUMMARY

The multiplication table of the quotient group C_{6}/C_{3} consists of two elements: C_{3} and ωC_{3}. The group C_{6} is generated by ω, with C_{3} defined as {1, ω^2, ω^4}. The multiplication table requires computing the products of the cosets: C_{3} × C_{3}, C_{3} × ωC_{3}, ωC_{3} × C_{3}, and ωC_{3} × ωC_{3}. This group is identified as a familiar cyclic group of order 2.

PREREQUISITES
  • Understanding of cyclic groups, specifically C_{n} notation
  • Familiarity with quotient groups and cosets
  • Basic knowledge of group multiplication
  • Experience with complex roots of unity, particularly ω
NEXT STEPS
  • Study the properties of cyclic groups and their generators
  • Learn about quotient groups and their significance in group theory
  • Explore the concept of cosets in more depth
  • Investigate the application of group theory in abstract algebra
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Students of abstract algebra, mathematicians interested in group theory, and anyone studying the properties of cyclic and quotient groups.

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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
 
Last edited:

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