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I want to classify the abelian groups of the form

\begin{align} \mathbb{Z}\times \mathbb{Z} \times \mathbb{Z} / \langle (n,n,n) \rangle \end{align} using fundamental theorem of finitely generated abelian groups.

From my previous post Classifying $ \frac{\mathbb{Z} \times\mathbb{Z}}{\langle(n,n)\rangle}$ via fundamental theorem of finitely generated abelian groups, I understand for $\mathbb{Z}\times \mathbb{Z} / \langle (n,n) \rangle \simeq \mathbb{Z} \times \mathbb{Z}_n$.

The easy way to see this is by considering its generators. $\mathbb{Z} \times \mathbb{Z} = \langle (1,1), (1,0) \rangle$ and moding out $n(1,1)$ I have the above expression.

Is the same procedure can apply for higher rank case? I mean generalization of $\mathbb{Z}^k$ mod by group generated by $n(1,\cdots, 1)$?

For $k=3$, $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} = \langle (1,1,1),(1,1,0),(1,0,1)\rangle $ and in this time I cannot mod $(1,1,1)$ freely....

Shaun
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phy_math
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1 Answers1

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As said in the comments, you can use the Smith normal form to get your result. However, we can simply notice that $e_1=(1,1,\ldots ,1),e_2=(0,1,0,\ldots,0),\ldots, e_k=(0,0,\ldots,0,1)$ is a basis of $\mathbb{Z}^k$ (easy: the base change matrix is upper triangular with determinant $1$).

Hence $\mathbb{Z}^k=\mathbb{Z}e_1\oplus\mathbb{Z}e_2\oplus\cdots\oplus\mathbb{Z}e_k$, while $\mathbb{Z}(n,n,\ldots,n)=n\mathbb{Z}e_1$.

The first isomorphism theorem applied to $f:\displaystyle\sum_{i=1}^k a_i e_i\in \mathbb{Z}^k\mapsto (\overline{a}_1,a_2,\ldots,a_k)\in\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}^{k-1}$ shows that your quotient group is isomorphic to $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}^{k-1}.$

GreginGre
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