Let $A$ be the Abelian group generated by $e,f,g$ such that:
\begin{eqnarray} 9e + 3f + 6g = 0 \\
3e + 3f + 0g = 0 \\
3e - 3f + 6g = 0
\end{eqnarray}
Determine the decomposition of $A$ as a direct sum of cyclic groups. I can diagonalize the relations matrix such that: $\begin{bmatrix} 3 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -1 & -1 \end{bmatrix} \cdot \begin{bmatrix} 12 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 3 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -1 & -1 \end{bmatrix}^{-1} = \begin{bmatrix} 9 & 3 & 6 \\ 3 & -3 & 0 \\ 3 & -3 & 6 \end{bmatrix}$. I believe this problem has something to do with a change of basis, but I do not know what to do. Is there a general procedure to solving this sort of problem?
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The general method is computing the Smith normal form of the matrix. There are many very similar questions, e.g.: 1, 2, 3 – Viktor Vaughn Jan 06 '20 at 15:25
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If you let $e'=e+f,f'=e+g$ then $<e',f',g>=<e,f,g>.$
The relations now are simply $3e'=6f'=0$ and so the group is the direct sum of $Z_3,Z_6$ and $Z$.
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You had the right idea of change of basis. In this case one sees that $3(e+f)=0$ is a great start and then use that to simplify the others by row operations. Subtract it from the top row, add it to the bottom row. – Jan 06 '20 at 00:25