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Let $H = \langle (6,2), (3,6)\rangle$, which is a subgroup of $\mathbb{Z}^2$ (denoted $H\leq \mathbb{Z}^2$). Show that $|\mathbb{Z}^2/H| = 30$ and that $\mathbb{Z}^2/ H$ is cyclic, and then find a surjective group homomorphism $\phi : \mathbb{Z}^2 \to \mathbb{Z}_{30}$ with $Ker(\phi) = H.$

I think that $\mathbb{Z}^2/ H = \{(r,s) + H : 0\leq r < 15, 0\leq s < 2\} =: T$ and $T = \langle (2,1) + H\rangle$. Indeed $(2,1) + H \in T$ and $T\leq \mathbb{Z}^2/H$ so $\langle (2,1) + H\rangle \subseteq T.$ Also, every element in $T$ is a multiple of $(2,1) + H$ so $T\subseteq \langle (2,1) + H\rangle$ (one can show this using the fact that $(6,2), (15,0)\in H$). Also, by repeated use of the division algorithm, one can show that every coset is in $T.$ To show they're distinct one can obtain a contradiction from assuming $(r_1, s_1) + H = (r_2, s_2) + H$ if $(r_1, s_1)\neq (r_2, s_2) $ . One can show $s_1 = s_2$ and if $r_1 \neq r_2$ then they must differ by a multiple of $15.$ Also, $\phi : \mathbb{Z}^2/ H \to \mathbb{Z}_{30}, \phi((r,s) + H) = s \cdot 15 + r$ is a group isomorphism, which shows $\mathbb{Z}^2/ H\cong Z_{30}$ and hence $|\mathbb{Z}^2/ H| = 30$.

However, I’m not sure how to find a surjective group homomorphism $f: \mathbb{Z}^2 \to \mathbb{Z}_{30}$ with $Ker(\phi) = H.$ I tried determining the values of $f(1,0)$ and $f(0,1)$ but apparently I seem to get a contradiction.

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  • $\begingroup$ Presumably you want to use the natural homomorphism with kernel $H$, and the question is to determine an element of $\mathbb{Z}^2$ mapping to a cyclic generator of the image? $\endgroup$
    – vujazzman
    Jan 17, 2021 at 0:51
  • $\begingroup$ In other words, you may want to construct a surjective group homomorphism $f:\mathbb{Z}^2\to\mathbb{Z}_{30}$ with $\ker(f)=H$. Then by the first isomorphism theorem, $\mathbb{Z}^2/H\cong\mathbb{Z}_{30}$, hence $\mathbb{Z}^2/H$ is cyclic of order $30$. $\endgroup$ Jan 17, 2021 at 0:55
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    $\begingroup$ Fix isomorphism $\varphi:\mathbb{Z}^2/H\to\mathbb{Z}_{30}$, and the quotient map $f:\mathbb{Z}^2\to\mathbb{Z}^2/H$. Then $\varphi\circ f:\mathbb{Z}^2\to\mathbb{Z}_{30}$ is what you needed. $\endgroup$
    – user632577
    Jan 17, 2021 at 0:57
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    $\begingroup$ Observe that $\det\pmatrix{6&2\\3&6}=30$. $\endgroup$
    – Berci
    Jan 17, 2021 at 1:38
  • $\begingroup$ Why is $\mathbb{Z}^2/ H = \{(r,s) + H : 0\leq r < 15, 0\leq s < 2\}$? Shouldn't that be $\mathbb{Z}^2/ H = \{(r,s) + H : r,s\in\Bbb Z\}$? $\endgroup$
    – user870827
    Jan 19, 2021 at 21:57

1 Answer 1

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Consider instead of the given generators (colomns in $A$ below) the ones in the columns of the following matrix $B$: $$ \underbrace{ \begin{bmatrix} 0 & 3 \\ -10 & 6 \end{bmatrix}}_{B} = \underbrace{ \begin{bmatrix} 6 & 3 \\ 2 & 6 \end{bmatrix}}_{\text{Given $A$}} \underbrace{ \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}}_T \ . $$ And note that $TAT=TB=\begin{bmatrix} 0 & 3 \\ -10 & 0 \end{bmatrix}$.

This suggest to define a surjective group morphism from the column vector with components $a,b\in\Bbb Z$ as follows (so that the columns of $B$, thus also those of $A$ are mapped to "zero"): $$ \begin{bmatrix} a\\ b \end{bmatrix} \to T \begin{bmatrix} a\\ b \end{bmatrix} = \begin{bmatrix} a\\ b-2a \end{bmatrix} =: \begin{bmatrix} a'\\ b' \end{bmatrix} \to \begin{bmatrix} a'\mod 3\\ b'\mod 10 \end{bmatrix} \in \overset{\displaystyle\Bbb Z/3}{\underset{\displaystyle\Bbb Z/10}\oplus} \cong \Bbb Z/30\ . $$

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