Counting Cosets in Abstract Algebra | Pinter's Self Study

  • #1
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Hi,
I am doing self study of Abstract Algebra from Pinter.
My doubt is regarding Chap 13 Counting Cosets:
A coset contains all products of the form "ah" where a belongs to G and h belongs to H where H is a subgroup of G. So each coset should contain the number of elements in H. Now the number of cosets should be the number of elements in G since each element of G is used to construct a coset. So the number of elements in all cosets should be number of elements in G*Number of elements in H - But the family of cosets is a partition of G and should have the same number of elements of G... there is definitely something wrong in the second-third line of this argument... but I am not able to pin it down
 
  • #2
Hi,
Now the number of cosets should be the number of elements in G since each element of G is used to construct a coset.

Do two different elements of G necessarily construct two different cosets ?

For example, suppose the two elements of [itex] G [/itex] are [itex] h1, h2 [/itex] and that [itex] h_1 \in H [/itex] and [itex] h_2 \in H [/itex].
 
  • #3
Thanks - The coset of h1 would be the set of elements h1*h and that of h2 would be h2*h for all h belonging to H - Not sure why they should not be different.
I am not able to understand why the number of cosets should not be equal to the number of elements in the set G. Am sure I'm missing something very basic.
 
  • #4
If [itex] h \in H [/itex] then the coset [itex] hH = H [/itex]. If you multiply an element of a subgroup by another element in the subgroup then the product is in the subgroup.
 
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Likes sdembi
  • #5
Understood - Thanks a ton...
 

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