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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

$R :=$ $$ \begin{bmatrix} -6 & 111 & -36 & 6\\ 5 & -672 & 210 & 74\\ 0 & -255 & 81 & 24\\ -7 & 255 &-81 & -10 \end{bmatrix} $$

  1. 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.

  2. 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.

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  • $\begingroup$ Why exactly do you think you choose $b_1=a_1+a_2$? There is a theorem that says that any finitely generated abelian group is isomorphic to a direct sum of cyclic groups (which may be finite or infinite), and 2) is asking you to find the orders of those cyclic groups. In fact 2) is easy once you have solved 1). To solve 1), first do unimodular tow operations on the matrix to bring it to upper triangular form - this corresponds to things like replacing one of your four equations by itself plus an integral multiple of another. After doing that, it should be easy to see what $b_1,\ldots,b_4$ are. $\endgroup$ – Derek Holt May 5 '14 at 19:26
  • $\begingroup$ @DerekHolt Okay so I've rearranged the matrix into the upper triangular form. $U :=$ $$ \begin{bmatrix} 1 & -37/2 & 6 & -1\\ 0 & 1 & -360/1159 & -158/1159\\ 0 & 0 & 1 & -6\\ 0 & 0 & 0 & 0 \end{bmatrix} $$ So is this where I get my bs from? Is it fine that I have got fractions or should I multiply everything by 1159? So $b_1 -\frac{37}{2} b_2 + 6b_3 - b_4 = 0$ $b_2 - \frac{360}{1159} b_3 - \frac{158}{1159} b_4 = 0$ $b_3 + 6b_4 = 0$ Is that right? How do I present my answer? $\endgroup$ – user113463 May 5 '14 at 20:00
  • $\begingroup$ No, you should be working over the integers, not over the rationals: there should be no fractions in the triangularized matrix. You can add an integral multiple of one row to another row, interchange two rows, or multiply a row by $-1$. $\endgroup$ – Derek Holt May 5 '14 at 21:09
  • $\begingroup$ Right okay, so one I have my matrix, do I then get my $b_i$s like that? Will that matrix be my D, in smith normal form? $\endgroup$ – user113463 May 5 '14 at 21:13
  • $\begingroup$ Well, if for example the triangularized matrix was $$\left[\begin{array}{cccc} 1&0&0&-2\\0&3&3&-24\\0&0&21&-126\\0&0&0&0 \end{array} \right],$$ then you would have $b_1=3b_2=21b_3=0$ with $b_1=a_1-2a_4$, $b_2=a_2+a_3-8a_4$,$b_3=a_3-6a_4$,$b_4=a_4$, and the group would be ${\mathbb Z}_3 \oplus {\mathbb Z}_{21} \oplus {\mathbb Z}$. $\endgroup$ – Derek Holt May 5 '14 at 22:05

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