Let \(G\) be the abelian group generated by \(x,y,z\) subject to \[ 15x+3y=0, \] \[ 3x+7y+4z=0, \] \[ 18x+14y+8z=0. \]
- Express \(G\) as a direct product of two cyclic groups.
- Express \(G\) as a direct product of cyclic groups of prime-power order.
- How many elements of \(G\) have order \(2\)?
Solution
Let \(C_m=\ZZ/m\ZZ\) and let \[ R= \begin{pmatrix} 15&3&0\\ 3&7&4\\ 18&14&8 \end{pmatrix} \] be the relation matrix with respect to the ordered generators \(x,y,z\).
1. The determinantal divisors of \(R\) are\[ \Delta_1=1, \qquad \Delta_2=12, \qquad \Delta_3=144. \]
The greatest common divisor of all entries of \(R\) is \(1\), because the entries include \(3\) and \(7\). Thus \(\Delta_1=1\).
The nonzero \(2\times2\) minors, up to sign, are \[ 96,\ 60,\ 12,\ 156,\ 120,\ 24,\ 84,\ 48. \] Their greatest common divisor is \(12\), so \(\Delta_2=12\).
Finally, \[ \det R = 15(7\cdot8-4\cdot14) -3(3\cdot8-4\cdot18) = 144. \] Hence \(\Delta_3=\abs{\det R}=144\).
2. The Smith invariant factors of \(R\) are\[ d_1=1, \qquad d_2=12, \qquad d_3=12. \]
For a full-rank integer matrix, the Smith invariant factors satisfy \[ \Delta_k=d_1d_2\cdots d_k. \] Using step 1, \[ d_1=\Delta_1=1, \qquad d_2=\frac{\Delta_2}{\Delta_1}=12, \qquad d_3=\frac{\Delta_3}{\Delta_2}=12. \]
3. For part 1,\[ \boxed{ G\cong C_{12}\times C_{12} }. \]
The presentation identifies \(G\) with the quotient of \(\ZZ^3\) by the subgroup generated by the rows of \(R\). By the Smith normal form theorem and step 2, \[ G \cong C_1\times C_{12}\times C_{12} \cong C_{12}\times C_{12}. \]
4. For part 2,\[ \boxed{ G\cong C_4\times C_4\times C_3\times C_3 }. \]
Since \(12=4\cdot3\) and \(\gcd(4,3)=1\), the Chinese remainder theorem gives \[ C_{12}\cong C_4\times C_3. \] Apply this to both factors in step 3 and reorder the direct factors.
5. For part 3, the number of elements of order \(2\) is\[ \boxed{3}. \]
In \(C_{12}\), the elements annihilated by \(2\) are exactly \(0\) and \(6\). Therefore the elements of \(C_{12}\times C_{12}\) whose order divides \(2\) are \[ (0,0),\qquad(6,0),\qquad(0,6),\qquad(6,6). \] The first has order \(1\), and each of the other three has order \(2\). Step 3 transfers this count to \(G\).
6. Q.E.D.
Steps 3, 4, and 5 answer parts 1, 2, and 3, respectively.