# 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 $G/H$ 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

$A=\begin{bmatrix} &&\\ &6&1\\ 3&&\\ 1&&1\\ \end{bmatrix}$

We can preform row operations by multiplying on the left by the matrix $P$ and column operations by multiplying on the right by the matrix $Q$

$P= \begin{bmatrix} &&&1\\ &1&& \\&3&1&-3\\ 1&&&\\ \end{bmatrix}$ ,

$Q=\begin{bmatrix} 1&6&-1\\ &1&\\ &-6&1\\ \end{bmatrix}$ to obtain the matrix

$PAQ=\begin{bmatrix} 1&&\\ &&1\\ &18&\\ &&\\ \end{bmatrix}$

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 '15 at 2:23
• I think I fixed it now, I'm looking to an explicit function, not just to identify the abelian group. – user135520 Nov 27 '15 at 1:14

## 1 Answer

This problem hinges on interpreting the Smith normal form of a matrix

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 $$PAQ = \begin{pmatrix} 1 & &\\ & 1 &\\ & & 18\\ & & \end{pmatrix}$$ 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 $18$.

• Thanks, I think this is as good as an answer as I could possible hope for. Thanks, again. – user135520 Nov 28 '15 at 15:26
• @Alphonse You've got a sharp eye! I've corrected my answer. – André 3000 Dec 11 '16 at 21:14
• @Alphonse : ahah, you're welcome :-). And +1, by the way! – Alphonse Dec 11 '16 at 21:14
• Does this method carry over to the case described here? math.stackexchange.com/questions/2856306/… – user437309 Jul 19 '18 at 5:59
• @user437309 Yes, see ancientmathematician's comment on your question. – André 3000 Jul 20 '18 at 2:32