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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

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  • $\begingroup$ See math.stackexchange.com/a/722055/589 for a similar but simpler example. $\endgroup$
    – lhf
    Nov 26, 2015 at 2:23
  • $\begingroup$ I think I fixed it now, I'm looking to an explicit function, not just to identify the abelian group. $\endgroup$
    – user135520
    Nov 27, 2015 at 1:14

1 Answer 1

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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 $$ PAQ = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 18\\ 0 & 0 & 0 \end{pmatrix} $$ 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 $18$.

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    $\begingroup$ Thanks, I think this is as good as an answer as I could possible hope for. Thanks, again. $\endgroup$
    – user135520
    Nov 28, 2015 at 15:26
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    $\begingroup$ Does this method carry over to the case described here? math.stackexchange.com/questions/2856306/… $\endgroup$
    – user557
    Jul 19, 2018 at 5:59
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    $\begingroup$ @user437309 Yes, see ancientmathematician's comment on your question. $\endgroup$ Jul 20, 2018 at 2:32
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    $\begingroup$ 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 $P^{-1} = \left(\begin{array}{rr} 4 & 1 \\ 1 & 0 \end{array}\right)$. $\endgroup$ Dec 3, 2021 at 3:27
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    $\begingroup$ 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 $q_2 \mapsto 6 r_2$. $\endgroup$ Dec 3, 2021 at 3:29

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