Finding Cosets with H = {0, 3, 6} under Z(9): Understanding the Starting Values

[semidevil]

so let H = { 0, 3, 6} under Z(9), and I need to find a + H.

the book shows

0 + H = 3 + H = 6 + H
1 + H = 4 + H = 7 + H
2 + H = 5 + H = 8 + H.

I"m not understanding why they start with 0, 1, 2. what gives that away?

 

[matt grime]

They find the coset of each element in the nice logical order that comes from the fact that 0<1<2, and that completely enumerates all the cosets. You've got to start somewhere, why not there?

 

[M98Ranger]

The starting values of 0, 1, and 2 are based on the definition of cosets. In this case, the cosets are defined as the set of all elements that can be obtained by adding an element from H to a given element in Z(9). Since H = {0, 3, 6}, the cosets will be of the form a+H, where a is any element in Z(9).

Starting with 0, 1, and 2 allows us to cover all possible elements in Z(9) and ensures that we do not miss any elements in the cosets. For example, if we started with a different set of values, such as 3, 4, and 5, we would not be able to represent the elements 0, 1, and 2 in the cosets.

Additionally, starting with 0, 1, and 2 allows us to easily identify the elements in each coset. For example, in the first coset (0+H), we can see that all elements in this coset will be multiples of 3, since 0+0=0, 0+3=3, and 0+6=6. Similarly, in the second coset (1+H), all elements will be 1 greater than a multiple of 3, and in the third coset (2+H), all elements will be 2 greater than a multiple of 3.

In summary, starting with 0, 1, and 2 allows us to have a systematic approach to finding all the elements in the cosets and ensures that we do not miss any elements.