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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 $C_3\times C_2$.

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    $\begingroup$ This is correct. And pretty much standard! Many a text may give a slight preference of putting the matrix into its Smith normal form, when you get to read the invariant factors as well (here $6,1,1$). Anyway, $C_6\simeq C_3\times C_2$, and all is well :-) $\endgroup$ – Jyrki Lahtonen Jul 2 '18 at 21:35
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This is perfect. You find $\left(\matrix{2& 1& 5\\-3& 0& 0\\ 0& 0& 2}\right)$ in two row operations, then $\left(\matrix{0& 1& 0\\-3& 0& 0\\ 0& 0& 2}\right)$ in two additional column operations.

This can then trivially be turned into $\left(\matrix{1& 0& 0\\0& 2& 0\\ 0& 0& 3}\right)$, which gives your result, but note that a few more operations would yield $$\left(\matrix{1& 0& 0\\0& 1& 0\\ 0& 0& 6}\right)$$ which is the Smith Normal Form of the original matrix. Of course, $C_6\simeq C_3\times C_2$.

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I think that's good, but you don't even need to go into $R$-modules. Since the elementary integer row operations correspond to group operations, you can derive the simpler set of generators from the first and deduce that the two sets of generators generate the same group.

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