# Finding an explicit isomorphism from $\mathbb{Z}^{4}/H$ to $\mathbb{Z} \oplus \mathbb{Z}/18\mathbb{Z}$

There was a past qualifying exam problem, I was having trouble with, it is stated below as follows:

In the group $$G= \mathbb{Z} \times \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}=\mathbb{Z}^{4}$$, let $$H$$ be the subgroup generated by $$[0,0,3,1], [0,6,0,0], [0,1,0,1]$$. Find an explicit isomorphism between $$G/H$$ and a product of cyclic groups.

I truly do not fully understand how to define such a map on $$G/H$$. I have given some thought to this.

We have that $$H$$ consists of integer combinations of the generators above, that is, if $$\xi \in H$$, then $$\xi=a[0,0,3,1]+b[0,6,0,0]+c[0,1,0,1]$$. In particular, we have that $$\xi$$ is in the image of a module homomorphism $$\mathbb{Z}^{3} \rightarrow \mathbb{Z}^{4}$$ whose matrix with respect to a standard basis $$(e_{1},e_{2}, e_{3})$$ and $$(f_{1}, f_{2}, f_{3}, f_{4})$$ for $$\mathbb{Z}^{3}$$ and $$\mathbb{Z}^{4}$$ respectively, will be $$A=\begin{bmatrix} 0&0&0\\ 0&6&1\\ 3&0&0\\ 1&0&1\\ \end{bmatrix}.$$

We can perform row operations by multiplying on the left by the matrix $$P$$ and column operations by multiplying on the right by the matrix $$Q$$ $$P= \begin{bmatrix} 0&0&0&1\\ 0&1&0&0\\ 0&3&1&-3\\ 1&0&0&0\\ \end{bmatrix},$$ $$Q=\begin{bmatrix} 1&6&-1\\ 0&1&0\\ 0&-6&1\\ \end{bmatrix}$$ to obtain the matrix $$PAQ=\begin{bmatrix} 1&0&0\\ 0&0&1\\ 0&18&0\\ 0&0&0\\ \end{bmatrix}.$$

I believe this tells us if $$\xi \in H$$, then $$\xi=[a,6b,18c,0]$$ for $$[a,b,c] \in \mathbb{Z}^{3}$$. From here, I think we can conclude that

$$\mathbb{Z}^{4}/H \cong \mathbb{Z} \times \mathbb{Z}/18\mathbb{Z}.$$

I do not see how to explicitly write a function between these two objects.

Do I think of $$\mathbb{Z}^{4}/H$$ as cosets?

The matrices $$P$$ and $$Q$$ above, tell us exactly the change of basis in the domain and codomain, I was wondering if I could use that.

Thanks

• See math.stackexchange.com/a/722055/589 for a similar but simpler example.
– lhf
Nov 26, 2015 at 2:23
• I think I fixed it now, I'm looking to an explicit function, not just to identify the abelian group. Nov 27, 2015 at 1:14

As you have said, we need to study the homomorphism $$\varphi: \mathbb{Z}^3 \to \mathbb{Z}^4$$ mapping the standard basis vectors to generators of the subgroup $$H$$. With respect to the standard bases $$\mathcal{E}$$ and $$\mathcal{F}$$ for $$\mathbb{Z}^3$$ and $$\mathbb{Z}^4$$, $$\varphi$$ has matrix $$A$$. We will find new bases $$\mathcal{B}$$ and $$\mathcal{C}$$ with respect to which $$\varphi$$ is represented by a (much simpler) diagonal matrix.
Computing the Smith normal form by row and column operations, I find that $$PAQ = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 18\\ 0 & 0 & 0 \end{pmatrix}$$ where $$P = \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0\\ 0 & -15 & 1 & -3\\ 1 & 0 & 0 & 0 \end{pmatrix} \qquad \text{and} \qquad Q = \begin{pmatrix} 1 & 5 & 6\\ 0 & 1 & 1\\ 0 & -5 & -6 \end{pmatrix} \, .$$ Note that $$P$$ and $$Q$$ both have determinant $$-1$$, hence are invertible over $$\mathbb{Z}$$. Interpreting $$P$$ and $$Q$$ as change of basis matrices, then $$PAQ = {_\mathcal{C} [\text{id}]_\mathcal{F}} \, {_\mathcal{F}[\varphi]_\mathcal{E}} \, {_\mathcal{E}[\text{id}]_\mathcal{B}}$$ for some bases $$\mathcal{B}$$ and $$\mathcal{C}$$ as above. Then $$\mathcal{B} = \left\{b_1 = e_1, b_2 = 5e_1 + e_2 - 5e_3, b_3 = 6e_1 + e_2 - 6e_3 \right\}$$, and since $${_\mathcal{F} [\text{id}]_\mathcal{C}} = P^{-1}= \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0\\ 3 & 15 & 1 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix}$$ we find $$\mathcal{C} = \{c_1 = 3f_3 + f_4, c_2 = f_2 + 15f_3, c_3 = f_3, c_4 = f_1\}$$. By the form of $$PAQ$$, then $$\varphi(b_1) = c_1$$, $$\varphi(b_2) = c_2$$, $$\varphi(b_3) = 18c_3$$ and $$\varphi(b_4) = 0$$, so $$H = \text{img}(\varphi) = \mathbb{Z}c_1 \oplus \mathbb{Z} c_2 \oplus \mathbb{Z}18 c_3$$. Thus \begin{align*} \frac{\mathbb{Z}^4}{H} = \frac{\mathbb{Z} c_1 \oplus \mathbb{Z} c_2 \oplus \mathbb{Z} c_3 \oplus \mathbb{Z} c_4}{\mathbb{Z}c_1 \oplus \mathbb{Z} c_2 \oplus \mathbb{Z}18 c_3} \cong \frac{\mathbb{Z}}{18\mathbb{Z}} \oplus \mathbb{Z} \, . \end{align*} More explicitly, the isomorphism is induced by the map $$\alpha_1 c_1 + \alpha_2 c_2 + \alpha_3 c_3 + \alpha_4 c_4 \mapsto (\overline{\alpha_3},\alpha_4)$$ where the bar indicates the residue mod $$18$$.
• In the example you give, with $A = \left(\begin{array}{rr} 4 & 6 \\ 1 & 3 \end{array}\right)$ , I get $P = \left(\begin{array}{rr} 0 & 1 \\ 1 & -4 \end{array}\right)$, $Q = \left(\begin{array}{rr} 1 & 3 \\ 0 & -1 \end{array}\right)$, and $D = \left(\begin{array}{rr} 1 & 0 \\ 0 & 6 \end{array}\right)$. Then $P^{-1} = \left(\begin{array}{rr} 4 & 1 \\ 1 & 0 \end{array}\right)$. Dec 3, 2021 at 3:27
• The columns $q_1, q_2$ of $Q$ form a basis for the domain of $A$; the quotient $(\mathbb Z \times \mathbb Z)/ \langle (4, 1), (6, 3) \rangle$ is the cokernel of left multiplication by $A$; the columns $r_1, r_2$ of $P^{-1}$ form a basis for the codomain of $A$, and left multiplication by $A$ sends $q_1 \mapsto r_1$ and $q_2 \mapsto 6 r_2$. Dec 3, 2021 at 3:29