I'm working on the matrix algorithm for finding the invariant factor decomposition of a free abelian group with n generators modded out by one of it's subgroups. I believe I can do the algorithm well however I was initially concerned that checking my solution wasn't obvious or any less difficult unless the subgroup is generated by one or maybe just two elements. I found a document from a UCB algebra class that claims to have a way to check by finding the new basis vector which defines the surjective homomorphism $\phi$ which maps our factor group to a new free group. And of course from the homomorphism theorem the Ker($\phi$) should include our subgroup. Here is the document
My only problem at this stage is that my linear algebra skills and experience are not so strong and I'm having a hard time following exactly how (and especially why) the change of basis for the new free abelian group is what it is given the matrix operations we make for the matrix representation of our subgroup. There is in fact a table that attempts to describe what change we make to our initial basis given each matrix operation but when I follow an example it does not seem to match up with the instructions in the table. Can anyone help by breaking down how exactly $(e_1, e_2, \ldots, e_n)$ becomes $(e_1', e_2', \ldots, e_n')$ and specifically how we know exactly in which way to write the later in terms of basis vectors from the former (ex. if we change $(e_1, e_2, e_3)$ to $(e_1 + 2e_2, e_2, e_3)$ did we add 2 times the first column to the second column or 2 times the 2nd column to the first column etc.)? Thank you!