Find, with proof, a product of cyclic groups isomorphic to \[ (\ZZ_{12}\times\ZZ_{12})/\langle(2,6)\rangle. \] Here \(\ZZ_n\) means \(\ZZ/n\ZZ\).
Solution
1. The required group is the cokernel of the homomorphism \(A:\ZZ^3\to\ZZ^2\) whose matrix is\[ A = \begin{pmatrix} 12&0&2\\ 0&12&6 \end{pmatrix}. \]
The group \(\ZZ_{12}\times\ZZ_{12}\) has the presentation \[ \ZZ^2/ \left\langle (12,0),(0,12) \right\rangle. \] Quotienting further by the class of \((2,6)\) adds the relation \((2,6)\). Thus the full relation subgroup is generated by \[ (12,0),\qquad (0,12),\qquad (2,6), \] which are exactly the three columns of \(A\).
2. Integer elementary row and column operations reduce \(A\) to\[ \begin{pmatrix} 2&0&0\\ 0&12&0 \end{pmatrix}. \]
Starting from \(A\), swap the first and third columns: \[ \begin{pmatrix} 12&0&2\\ 0&12&6 \end{pmatrix} \longmapsto \begin{pmatrix} 2&0&12\\ 6&12&0 \end{pmatrix}. \] Then replace row \(2\) by row \(2-3\) times row \(1\): \[ \begin{pmatrix} 2&0&12\\ 6&12&0 \end{pmatrix} \longmapsto \begin{pmatrix} 2&0&12\\ 0&12&-36 \end{pmatrix}. \] Next replace column \(3\) by column \(3-6\) times column \(1\): \[ \begin{pmatrix} 2&0&12\\ 0&12&-36 \end{pmatrix} \longmapsto \begin{pmatrix} 2&0&0\\ 0&12&-36 \end{pmatrix}. \] Finally replace column \(3\) by column \(3+3\) times column \(2\): \[ \begin{pmatrix} 2&0&0\\ 0&12&-36 \end{pmatrix} \longmapsto \begin{pmatrix} 2&0&0\\ 0&12&0 \end{pmatrix}. \]
Column operations multiply \(A\) on the right by an element of \(\GL_3(\ZZ)\), so they do not change the subgroup generated by the columns. Row operations multiply on the left by an element of \(\GL_2(\ZZ)\), hence apply an automorphism of the ambient group \(\ZZ^2\) and therefore preserve the cokernel up to isomorphism.
3. The cokernel of the reduced matrix is\[ \ZZ_2\times\ZZ_{12}. \]
The reduced relation subgroup is generated by \((2,0)\) and \((0,12)\). Therefore \[ \ZZ^2/ \left\langle (2,0),(0,12) \right\rangle \cong \ZZ/2\ZZ\times\ZZ/12\ZZ = \ZZ_2\times\ZZ_{12}. \]
4. Hence\[ \boxed{ (\ZZ_{12}\times\ZZ_{12})/\langle(2,6)\rangle \cong \ZZ_2\times\ZZ_{12} }. \]
Combine steps <1>1–<1>3.
5. Q.E.D.
Step 4 gives the requested product of cyclic groups.