Linear algebra - Quotient Group

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Homework Statement

Let [tex]G[/tex] be subgroup of [tex]Z^2=Z \times Z[/tex] spanned by [tex](4,2), (6,-12)[/tex]. Compute quotient [tex]Z^2/G[/tex].

Homework Equations

The Attempt at a Solution

I was under the impression that I am supposed to write the vectors as columns in a matrix, and then compute the Smith-Normal form. Doing so, I obtain a 2x2 identity matrix, and I am not sure how to interpret the two 1's. Thanks in advance.

 

[mighty2000]

Hi there!

You are correct that you should write the vectors as columns in a matrix. This is called the "generating set" of the subgroup G. In this case, we have G = <(4,2), (6,-12)>.

To compute the quotient Z^2/G, we need to find all the cosets of G in Z^2. A coset is a set of the form g + G = {g + x | x ∈ G}, where g is some element of Z^2.

In this case, our cosets will look like (a,b) + G, where (a,b) is some element of Z^2. To find all the cosets, we need to find all the possible values of (a,b) in Z^2.

To do this, we can use the fact that Z^2 is isomorphic to Z × Z, and we can think of (a,b) as (a,0) + (0,b). So we can find all the possible values of (a,0) and (0,b) separately.

For (a,0), we have the possible values of a as 0, 1, 2, 3, 4, 5, ... and for (0,b), we have the possible values of b as 0, 1, -2, -3, -4, -5, ...

So our cosets will look like (0,0) + G, (1,0) + G, (2,0) + G, (3,0) + G, (4,0) + G, (5,0) + G, ... and (0,0) + G, (0,1) + G, (0,2) + G, (0,3) + G, (0,4) + G, (0,5) + G, ...

Now, we can compute these cosets by adding the elements of G to each of these possible values. For example, for (1,0) + G, we have (1,0) + (4,2) = (5,2) and (1,0) + (6,-12) = (7,-12). So this coset is {(5,2), (7,-12)}.

Continuing in this way, we can find