How many distinct H cosets are there?

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Homework Help Overview

The discussion revolves around the cyclic group Cn = of order n and the subgroup H= where m divides n. Participants are exploring how to determine the number of distinct H cosets and how to describe these cosets explicitly.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to clarify the interpretation of the subgroup H and its elements. Questions are raised about when elements of the form gmk are equal for different integers k, and how to express the conditions under which these elements coincide.

Discussion Status

The discussion is active, with participants engaging in back-and-forth questioning to refine their understanding of the relationships between the elements of the group and subgroup. Some guidance is being offered regarding the implications of the cyclic nature of the group and the order of elements.

Contextual Notes

Participants are considering the implications of Lagrange's Theorem and the specific values of k in relation to the order of the group and subgroup. There is an emphasis on understanding the distinct values generated by the subgroup as k varies.

Kate2010
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Homework Statement



Consider the cyclic group Cn = <g> of order n and let H=<gm> where m|n.

How many distinct H cosets are there? Describe these cosets explicitly.

Homework Equations



Lagrange's Theorem: |G| = |H| x number of distinct H cosets

The Attempt at a Solution



|G| = n
I'm unsure if I have interpreted H correctly, I think it is the cyclic group generated by gm so contains gmk for integers k. I don't know how I can work out the order of H from this, or how to describe the cosets.
 
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Fine so far. When will gmk=gmk' for different integers k, k'?
 
Would it be when k' is a multiple of k?
 
No. If you have a cyclic group (g) of order 20, and H=(g2) then the values g2x0, g2x1, g2x2, ... g2x9 will all be different, then you start again. Here m is 2 and k goes from 0 to 9, but for instance 1 divides each value of k, and 2 divides 4, but g2 is not the same as g4,g 6,g8 etc. and g2x2=g4 is not the same as g2x4=g8.

How would you get the powers of g in the equation gmk=gmk' to the same side of the equals sign? Then you can use gn=e.
 
Actually if you think about it, if gmk=gmk' were equivalent to k' is a multiple of k, it would also have to be equivalent to k is a multiple of k'.

This would mean the two values would be the same iff k=±k', but that would imply an infinite set ogf different values and there are only n.
 
Is it when k=n/m?
 
Well I think it's getting closer. Would that work with the example I gave? n/m in that instance would be 10. But where does k' come into it?

We want to know how many different values (gm)k runs through as k runs through all the integers. It can be at most n, but will it be all n? Would (g2)k give all 20 values in the example. Would you get g5 for some k for instance?

The idea of trying to find when two different k give the same result is that you would probably then see how many different groups of k give the same result. The size of H is then the number of groups. But if you look at the example carefully the result will probably become pretty obvious.

What for instance would be the next value of k in the example such that (g2)k=(g2)3?
 

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