Finding suitable basis for a free abelian finitely generated group.

Question

I am stuck with this exercise forever... I was barely taught about it, English is not my mother language and in any other phrasing it is not coherent with my material.I'd really appreciate your help. What I know is that $\{e_1,e_2\}$ is a basis of a free, finitely generated abelian group. There is a subgroup $B$ generated by $\{2e_1+3e_2,2e_1+6e_2\}$. I have to find a basis ${f_1,f_2}$ and positive integer $d_1|d_2|\cdots|d_r$ such that $\{d_1f_1, \cdots ,d_rf_r\}$ is a basis of $B$. In an example a teacher gave us, with other data(I will explain what he did with the data of this exercise instead.), he built a homomorphism from $f:\mathbb{Z}^2 \rightarrow A$, such that $f(1,0)=2e_1+3e_2, f(0,1)=2e_1+6e_2$. Im$f$ is of course $B$. Then classifying the columns of the matrix that represents the homomorphism according to $\{e_1,e_2\}$. In his example it would work well, but not here. I am frustrated. I'd like your help. Thanks :)

Answer 1

You can represent $f$ as a matrix, and apply Smith's algorithm, which is essentially an analogue of Gaussian elimination. My teacher got me used to writing matrices by rows, so I will represent $f$ like so:

$ f =\begin{bmatrix}2 & 3\\2 & 6\end{bmatrix} \to \begin{bmatrix}2 & 3\\0 & 3 \end{bmatrix} \to \begin{bmatrix}2 & 0\\0 & 3\end{bmatrix}$

In each step, I subtracted one row from another. The first arrow is basically the statement: "if $\{2e_1+3e_2,2e_1+6e_2\}$ generates $B$, then so does $\{2e_1+3e_2,3e_2\}$." If you can understand why that is true, you will be well on your way to answering your problem :) .

EDIT: I'll leave this since someone thought it was useful, but it does not fully answer the question.