Form an abelian group \(M\) as the quotient of \(\mathbb{Z}^3\) by the elements \((0,4,2)\), \((-1,-4,-1)\), and \((0,0,-2)\). What abelian group is \(M\)? (That is, find its isomorphism type).
Solution
Let \(A\) be the integer matrix whose rows are the three relations: \[ A = \begin{pmatrix} 0 & 4 & 2 \\ -1 & -4 & -1 \\ 0 & 0 & -2 \end{pmatrix}. \] Then \(M=\ZZ^3/\ZZ^3A\), where \(\ZZ^3A\) is the row space of \(A\).
1. The Smith normal form of \(A\) is \(\operatorname{diag}(1, 2, 4)\).
Integer row operations replace the relations by another generating set of the same subgroup, and invertible integer column operations change the basis of \(\ZZ^3\); neither changes the isomorphism type of \(M\). Subtract \(4\) times the first column from the second and the first column from the third, then negate the second row. The rows become \((0,4,2)\), \((1,0,0)\), \((0,0,-2)\). The remaining \(2\times2\) block in the last two columns is \(\begin{pmatrix}4&2\\0&-2\end{pmatrix}\). Subtracting twice its second column from its first gives \(\begin{pmatrix}0&2\\4&-2\end{pmatrix}\); adding its first row to its second gives \(\begin{pmatrix}0&2\\4&0\end{pmatrix}\); swapping its columns gives \(\operatorname{diag}(2,4)\). Since \(1\mid2\mid4\), this is the Smith normal form. As a check, \(\abs{\det A}=8=1\cdot2\cdot4\).
2. \(M \cong \boxed{\ZZ/2 \oplus \ZZ/4}\).
By step 1, \(M\cong\ZZ/1\oplus\ZZ/2\oplus\ZZ/4\), and \(\ZZ/1=0\).
3. Q.E.D.
Step 2 gives the isomorphism type.