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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

user26857
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user135520
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1 Answers1

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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 $18$.

user26857
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Viktor Vaughn
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    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
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    Does this method carry over to the case described here? https://math.stackexchange.com/questions/2856306/how-to-find-mathbb-z-12-times-mathbb-z-12-a4-a6 – user557 Jul 19 '18 at 05:59
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    @user437309 Yes, see ancientmathematician's comment on your question. – Viktor Vaughn Jul 20 '18 at 02:32
  • In general, is there a way to see what the isomorphism is? For instance, the Smith Normal Form tells us that $(\mathbb Z \times \mathbb Z)/ \langle (4, 1), (6, 3) \rangle \cong \mathbb Z / 6 \mathbb Z,$ but one of the factors is zero, hence the isomorphism is more complicated. – Dylan C. Beck Dec 02 '21 at 01:22
  • @DylanC.Beck I don't understand---isn't that exactly what the last part of my answer shows how to do? Since the first two columns of $PAQ$ have $1$s on the diagonal, those two factors, generated by $c_1$ and $c_2$, are killed and become $0$ in the quotient, while $c_3$ generates the $\mathbb{Z}/18\mathbb{Z}$ factor and $c_4$ generates the $\mathbb{Z}$ factor. – Viktor Vaughn Dec 03 '21 at 03:24
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    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 $P^{-1} = \left(\begin{array}{rr} 4 & 1 \ 1 & 0 \end{array}\right)$. – Viktor Vaughn Dec 03 '21 at 03:27
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    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 $q_2 \mapsto 6 r_2$. – Viktor Vaughn Dec 03 '21 at 03:29
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    Thus the isomorphism $(\mathbb Z \times \mathbb Z)/ \langle (4, 1), (6, 3) \rangle \overset{\sim}{\to} \mathbb Z / 6 \mathbb Z$ is given by $r_1 = \begin{pmatrix} 4 \ 1 \end{pmatrix} \mapsto \overline{0}$ and $r_2 = \begin{pmatrix} 1 \ 0 \end{pmatrix} \mapsto \overline{1}$. – Viktor Vaughn Dec 03 '21 at 03:29
  • I'm not sure I see what you mean. How can $r_1 = (4, 1)^t$ be a basis element of $(\mathbb Z \times \mathbb Z) / \langle (4, 1), (6, 3) \rangle$? It is zero. I suppose the point is that $(1, 0)^t \mapsto \bar 1$ and $(0, 1)^t \mapsto \bar 2,$ but then, the kernel consists of the column vectors $(a, b)^t$ with $6 \mid (a + 2b),$ and it's unclear why this is zero. – Dylan C. Beck Dec 09 '21 at 19:51
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    No, $(4,1)^T$ and $(1,0)^T$ form a basis for the codomain of multiplication by $A$, which is $\mathbb{Z} \times \mathbb{Z}$. The map $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}/6\mathbb{Z}$ given by $\begin{pmatrix} 4 \ 1 \end{pmatrix} \mapsto \overline{0}$ and $\begin{pmatrix} 1 \ 0 \end{pmatrix} \mapsto \overline{1}$ is surjective with kernel $\langle (4, 1), (6, 3) \rangle$ and then one can apply the First Isomorphism Theorem. – Viktor Vaughn Dec 09 '21 at 23:13
  • Could one instead use the columns of $P$ to define the isomorphism? It makes more sense to me that the kernel of this map is $A \mathbb Z^m$ because $P(A \mathbb Z^m) = PAQ \mathbb Z^m,$ and the latter is zero in the quotient. – Dylan C. Beck Dec 21 '21 at 17:22