Determine the isomorphism class of M/T(M) 
Let $M=\Bbb{Z}\oplus\Bbb{Z}\oplus\Bbb{Z}$ and $T: M\rightarrow M$ given by $T(x,y,z)=(4x+2z,2y,2x+10z)$. Show the cokernel $M/T(M)$ is an abelian group of order $72$, and determine its isomorphism class.  

I tried to solve it by looking at when the linear system has solution over Z, but it didn't get me anywhere.
I feel like this should not be an awefully hard question, but I don't know how to start. Can you give me some help? Thank you very much. 
 A: $T(M)$ has generators $(4,0,2)$, $(0,2,0)$, $(2,0,10)$, so we have the relations matrix 
$$\left(
  \begin{array}{ccc}
    4 & 0 & 2 \\
    0 & 2 & 0 \\
    2& 0 & 10 \\
  \end{array}
\right)$$
Computing the Smith normal form for this matrix gives
$$\left(
  \begin{array}{ccc}
    2 & 0 & 0 \\
    0 & 2 & 0 \\
    0& 0 & 18 \\
  \end{array}
\right)$$
and hence $M/T(M) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/18$ which has order $72$.
A: If you try to solve $Ax=b$ where
$A= \left(
  \begin{array}{ccc}
    4 & 0 & 2 \\
    0 & 2 & 0 \\
    2& 0 & 10 \\
  \end{array}
\right)$
 and  $b=\left(\begin{array}{c}b_{1}\\b_{2}\\b_{3}\end{array}\right)$
you need check which condition $b_{1}, b_{2}$ and $b_{3}$ must have in order to have a solution in $\mathbb{Z}^{3}$.
In this case $T(M)=36\mathbb{Z}\oplus 2\mathbb{Z}\oplus9\mathbb{Z}$
So
$\begin{eqnarray}\frac{M}{T(M)}&=&\frac{\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}}{36\mathbb{Z}\oplus 2\mathbb{Z}\oplus9\mathbb{Z}}\\
&\cong&\frac{\mathbb{Z}}{36\mathbb{Z}}\oplus\frac{\mathbb{Z}}{2\mathbb{Z}}\oplus\frac{\mathbb{Z}}{9\mathbb{Z}}\\
&\cong& \mathbb{Z}_{36}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{9} \end{eqnarray}$
I hope it helps you.
