Determine the rank and the elementary divisors of the following group: $A/H$ with $A \subset \mathbb{Z}^5$ the group of all $5-$tuples with sum $0$ and $H = A \cap B(\mathbb{Z}^5)$ where $$ B=\begin{bmatrix} -13& 1 & 1 & 0 & 0 \\ 1& -13 & 1 & 0 & 0 \\ 1&1&-1&1&1\\ 0&0&1&-2&0\\ 0&0&1&0&-3\\ \end{bmatrix} $$ I'm having trouble understanding how to go about doing this, my solution so far is given as follows:
Using the technique of repeatedly projecting the last element of any element in $A$, we find $a=(0,0,0-1,1) \in A$ so $\pi_5(a)$=$1$ so $\pi_5(A)=\mathbb{Z}$. Continuing like this we obtain $A$ in matrix form $$ \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ -1&-1&-1&-1\\ \end{bmatrix} $$ We want to find $A/(A\cap B(\mathbb{Z}^5))$. Using the theorem: Given a group $G$ and, $H \subset G$ a subgroup, and a normal subgroup $N \subset G$. Then $H/(H \cap N) \cong HN/N$. Since all subgroups of abelian groups are normal then $A/(A\cap B(\mathbb{Z}^5)) \cong AB(\mathbb{Z}^5)/A$. Then if we use the algorithm to change the bases of the groups represented by $A$ and $B$ we see that $A$ is represented by $$ \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 0&0&0&0\\ \end{bmatrix} $$ and $B$ is represented by $$ \begin{bmatrix} 7&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&0\\ \end{bmatrix} $$ So $A \cong \mathbb{Z}^4$ and $B \cong \mathbb{Z}^3 \times \mathbb{Z}/7\mathbb{Z}$. So $A/(A\cap B(\mathbb{Z}^5)) \cong AB/B \cong (\mathbb{Z}^7 \times \mathbb{Z}/7\mathbb{Z})/\mathbb{Z}^4 \cong \mathbb{Z}^{11} \times \mathbb{Z}/7\mathbb{Z}$. Rank $11$ and elementary divisor $7$.
But I'm not sure if I'm correctly interpreting $AB(\mathbb{Z}^5)/A$. I know that this is the quotient group of the group $AB(\mathbb{Z}^5)$ (which is the group whose elements are sums of 5-tuples which are linear combinations of the columns of $A$ and $B$). How would I then take the quotient group with $A$?
