Presentation of abelian groups and classification. One of my friends today asked me a question. I am unable to see a clear way to attack such a problem. Here it is:
Let $G$ be a finite abelian group with generators $a,b,c,d$ and relations : $2a = 4b+c,4c =d-2b\ \text{and}\ a+b+c+d = 0$. Now we wish to write $G$ as a product of its cyclic subgroup\classify it.
Now form the given information, I can see that $G$ can be finitely presented. And after solving the equations, I was able to reduce the relations and obtain the following:
$G = \langle a,b,c,d \ |\ 17b =11 a, 17c = -10a, 17d =-18a \rangle.$
Now how do I proceed from here, I thought of calculating the order but I don't know how to do so for a given presentation. In particular, is there some algorithm\formula to calculate\classify finitely presented abelian groups?
Maybe it can be done by using techniques for finitely generated abelian groups (invariant forms etc??) Please provide hints.
Thank you.
 A: I'm going to guess that what you did so far was to use "invertible row operations" to rewrite the relations. For each such operation, one takes a pair of distinct relations $R_i,R_j$ and some integer $m$, and one replaces the relator $R_i$ by the relator $R_i - m R_j$. And you chose these operations in a careful sequence to get a matrix which is in some kind of normal form.
If this is indeed what you did then good so far. Notice that the generating set does not change under these row operations. Let me rewrite the generators and relators in your second presentation in homogeneous form
$$17 b - 11 a = 0
$$
$$17 c + 10 a = 0
$$
$$17 d + 18 a = 0
$$
and then let me write these coefficients as a matrix, whose columns are headed, in order, by $b,c,d,a$:
\begin{pmatrix}
17 & 0 & 0 & -11 \\
0 & 17 & 0 & 10 \\
0 & 0 & 17 & 18
\end{pmatrix}
Besides row operations, you can use column operations to rewrite the presentation by replacing certain generators with certain other generators. For example, suppose that you let $b = b'+a$ and use that to remove the relator $b$. The equation
$$17 b - 11 a = 0
$$
becomes
$$17 (b'+a) - 11 a = 0
$$
or
$$17 b' + 6a = 0
$$
This affects the first row of the matrix but not the second or third, and one obtains
a new matrix headed by columns labelled $b',c,d,a$:
\begin{pmatrix}
17 & 0 & 0 & 6 \\
0 & 17 & 0 & 10 \\
0 & 0 & 17 & 18
\end{pmatrix}
In other words, the invertible column operation "add column 1 to column 4" is equivalent to the invertible generator replacement rule "replace generator 4 by generator 1 plus generator 4", and you can be therefore guaranteed that the exact same group is being presented, but now with a simpler matrix.
So, your question comes down to getting the matrix into the simplest form that you can by invertible row and column operations. And let me be explicit that this means operations which are invertible over the integers.
Perhaps you can take it from here (but with a hint: the Euclidean algorithm for finding g.c.d.'s is important).
A: Let $g=2a-3b$. Let $H$ be the subgroup of $G$ generated by $g$. 
Then from $11a-17b=0$, 
$$ 6g=12a-18b=a-b\in H,$$
hence $a=3(a-b)-g=5g\in H$, $b=a-(a-b)=5g-6g=-g\in H$, as well as $c=2a-4b=14g\in H$, and $d=2b+4c=54g\in H$. In other words $G=H$ is cyclic.
The last relation tells us that
$$ 0=a+b+c+d=5g-g+14g+54g=72g.$$
We verify that we obtain a homomorphism $G\to\Bbb Z/72\Bbb Z$ by sending $$a\mapsto 5,\quad b\mapsto -1,\quad c\mapsto 14,\quad d\mapsto 54.$$
