Invariant factor decomposition of quotient group of two subgroups of $\mathbb{Z}^n$.

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 $A$?

• I have not tried to check your calculation, but your answer makes no sense. The ranks of subquotients of ${\mathbb Z}^5$ are at most $5$, so the ${\mathbb Z}^7$ cannot be correct. In fact $A$ is free abelian of rank $4$, so the answer must have rank at most $4$. Jan 14, 2018 at 12:53
• Note also that $B$ is a subgroup of a free abelian group, so must itself be free abelian. In fact $B \cong {\mathbb Z}^4$. Also $AB \cong {\mathbb Z}^5$, so the answer must be ${\mathbb Z} \times K$ for some finite abelian group $K$. Since I am lazy, I just did the calculation in Magma, and the answer is ${\mathbb Z} \times {\mathbb Z}/7{\mathbb Z}$, so your answer is close - you just have the free rank wrong. Jan 14, 2018 at 13:39
• @DerekHolt thanks, I see that what I did makes no sense, but I think I must have gotten $\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z}$ by coincidence because my process was different. I see that $B \cong \mathbb{Z}^4$ I don't know where I got $\mathbb{Z}^3 \times \mathbb{Z}/7\mathbb{Z}$ from. So did you find the decomposition for the matrix $A*B$? like actual matrix multiplication? In such a case what does $AB/B$ mean? Jan 14, 2018 at 14:01
• I am going to write $Z$ for ${\mathbb Z}$. $A$ and $B$ are both subspaces of $Z^5$. Since you are treating $Z^5$ as an additive group here it would make more sense to write $A + B$ rather than $AB$, and then $(A+B)/B \cong A/A \cap B$. In fact $A+B = Z^5$, so the question is equivalent to computing $Z^5/B$, which is much easier. The matrix that you computed that represents $B$ is what is called the Smith Normal Form of $B$. From that, it is easy to see that $Z^5/B \cong Z + Z/7Z$. Jan 14, 2018 at 19:57
• No you calculated the Smith Normal Form correctly. A $0$ on the diagonal denotes an infinite factor, and a number $k$ denotes a factor $Z/kZ$. So the answer is $Z \times Z/7Z \times (Z/Z)^3 \cong Z \times Z/7Z$. Jan 15, 2018 at 9:46