Let $A_R:= \langle a_1, a_2, a_3, a_4 | R \circ \underline{a} = 0\rangle $ be the finitely generated abelian group, determined by the relation-matrix
$R :=$ $$ \begin{bmatrix} -6 & 111 & -36 & 6\\ 5 & -672 & 210 & 74\\ 0 & -255 & 81 & 24\\ -7 & 255 &-81 & -10 \end{bmatrix} $$
Find new generators $b_i$ such that $A_R \cong A_D := \langle b_1, b_2, b_3, b_4 | D \circ \underline{b} = 0\rangle$ with $D$ in the smith normal form.
Determine the isomorphism type of $A_R$.
I have a vague idea of what to do here. I think I choose $b_1 = a_1 + a_2$ , $b_2 = a_2 + a_3, b_3 = a_3 + a_4$ and $b_4 = a_4$ bt that's as far as I get before the example I managed to find off the internet becomes quite different.
Is the next step to set all the rows = 0 in terms of a? i.e
$-6 a_1 + 111a_2 -36a_3 + 6a_4 = 0$
$5 a_1 -672a_2 + 210a_3 + 74a_4 = 0$
$0 a_1 - 255a_2 + 81a_3 + 24a_4 = 0$
$-7 a_1 + 255a_2 - 81a_3 - 10a_4 = 0$
and then... what? From here on I have not only no idea what to do but such a finite grasp on the abstract idea of what I'm trying to do here that I have no real hope of figuring it out alone.
What am I even aiming for? I know the smith normal form is a diagonal matrix but that's about as much as I know from that.
As for 2), absolutely no idea what that means.
Sorry for seeming slow. I'm trying to teach myself maths in a week.