Cayley table of a group

  • #1
Just by looking at the cayley table of a group and looking at its subgroups, is their a theorem or something which tells you if the right and left cosets are equal?

I have question to do and I would love to half the workload by not having to to work out the same thing twice.
Thanks
 

Answers and Replies

  • #4
When multiplying cosets, can you just quite simply multiply them together?
Our teacher said something about their could be a potential problem?
What could the problem be?
 
  • #5
jbunniii
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When multiplying cosets, can you just quite simply multiply them together?
Our teacher said something about their could be a potential problem?
What could the problem be?
You can always multiply cosets, but the result is not necessarily a coset. In other words, the set of right or left cosets is not generally a group. Thus you can form the product ##aHbH##, which is the set of all elements of the form ##ah_1bh_2##, but this does not generally equal ##abH##, nor can it generally be written in the form ##gH## at all.

However, if ##H## is a normal subgroup, then the left and right cosets are the same (##aH = Ha##), so we just call them cosets, and the set of cosets does form a group. In this case, the product ##aHbH## can be simplified as follows:
$$aHbH = a(Hb)H = a(bH)H = abHH = abH$$
 
  • #6
You can always multiply cosets, but the result is not necessarily a coset. In other words, the set of right or left cosets is not generally a group. Thus you can form the product ##aHbH##, which is the set of all elements of the form ##ah_1bh_2##, but this does not generally equal ##abH##, nor can it generally be written in the form ##gH## at all.

However, if ##H## is a normal subgroup, then the left and right cosets are the same (##aH = Ha##), so we just call them cosets, and the set of cosets does form a group. In this case, the product ##aHbH## can be simplified as follows:
$$aHbH = a(Hb)H = a(bH)H = abHH = abH$$
Thank you for the good explanation.
 

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