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Say I wanted to determine the structure of an abelian group, but the matrix I get when computing the Smith Normal Form is not square.

For a simple example, consider $G=\mathbb{Z}^2/\langle (9,6)\rangle$. Then finding the SNF: $$ \begin{pmatrix} 9\\ 6 \end{pmatrix} \sim \begin{pmatrix} 3\\ 6 \end{pmatrix} \sim \begin{pmatrix} 3\\ 0 \end{pmatrix} $$ I think this means $\mathbb{Z}/(3)$ is an invariant factor. Does this just mean $G\simeq\mathbb{Z}\oplus\mathbb{Z}/(3)$, where I have to add another copy of $\mathbb{Z}$ to account for the "missing" column?

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Yes, if $G = {\mathbb Z}^m/\langle v_1,\ldots, v_n \rangle$ with $n<m$, then the $m-n$ missing columns in your SNF all give you a factor ${\mathbb Z}$. So the result is ${\mathbb Z}^{m-n} \oplus A$, where $A$ is what you get from the columns of the SNF matrix. If $n>m$, then you just ignore the rightmost $n-m$ columns of the SNF, which will all be zero columns.

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