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The full question is: Find the rank and elementary divisors (invariant factors) of $A/H$ with $A \subset \mathbb{Z}^5$ the group of all 5-tuples with sum $0$ and $H = A \cap B(\mathbb{Z}^5)$.

With $B$ the matrix \begin{bmatrix} -13& 1 & 1 & 0 & 0 \\ 1& -13 & 1 & 0 & 0 \\ 1&1&-1&1&1\\ 0&0&1&-2&0\\ 0&0&1&0&-3\\ \end{bmatrix}

Now my understanding is that we have to use the fundamental theorem of finitely generated abelian groups:

There exists a unique integer $r \geq 0$ and a unique finite sequence $(d_{1},...,d_{m})$ with all $d_{i} \in \mathbb{Z}$ and $d_{i}$ and $d_{m}|d_{m-1}|...|d_{1}$ such that $A \cong \mathbb{Z}^r \times \mathbb{Z}/d_{1}\mathbb{Z} \times ... \times \mathbb{Z}/d_{m}\mathbb{Z}$.

But I'm not even sure if I understand the question fully, or the notation. What does $B(\mathbb{Z}^5)$ even mean in this case? Is it $B$ multiplied by a $5\times1$ vector (tuple)? Or does it mean something else?

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$B(\mathbf Z^5)$ seems to denote the image of $\mathbf Z^5$ by the endomorphism associated to the matrix $B$. It is the submodule of $\mathbf Z^5$ generated by the column vectors of the matrix. So you're asked to find a basis of $\mathbf Z^5$ and a basis of the intersection of the submodules which satisfy the conditions of the fundamental theorem of finitely generated modules over a P.I.D.

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  • $\begingroup$ Thank you! I think I can piece it together. $\endgroup$
    – Daniel
    Commented Jan 6, 2018 at 13:40

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