Consider the quotient of $\mathbb Z^3$ by the subgroup generated by $(2,1,5),(1,2,10),(2,1,7)$. Write it as a product of cyclic groups.
I was wondering if this solution is complete and rigorous enough?
Recall that any homomorphism of $R$-modules $R^n\to R^m$ is given by a matrix $A$ with entries in $R$, and we say that $A$ is a presentation matrix of the quotient module $R^m/AR^n$.
In our case $m=n=3$, $R=\mathbb Z$. Let $A$ be the matrix whose columns are $(2,1,5)^t,(1,2,10)^t,(2,1,7)^t$. Then $AR^3$ is the subgroup of $R^3$ from the question. $A$ is a presentation matrix for the quotient group we need to identify. After using elementary integer row and column operations, the matrix reduces to the matrix with columns $(1,0,0)^t$, $(0,3,0)^t$, $(0,0,2)^t$. Since these operations yield a matrix that present the same module, we see that the quotient group is isomorphic to $C_3\times C_2$.