Why does counting cosets seem to give G times H elements?

[sdembi]

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

 

[Stephen Tashi]

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 $G$ are $h1, h2$ and that $h_1 \in H$ and $h_2 \in H$.

 

[sdembi]

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.

 

[Stephen Tashi]

If $h \in H$ then the coset $hH = H$. If you multiply an element of a subgroup by another element in the subgroup then the product is in the subgroup.

 

[sdembi]

Understood - Thanks a ton...