surjective group homomorphism between $\mathbb{Z}^2$ and $\mathbb{Z}_{30}$

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.

• 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? Commented Jan 17, 2021 at 0:51
• 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 $30$. Commented Jan 17, 2021 at 0:55
• 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.
– user632577
Commented Jan 17, 2021 at 0:57
• Observe that $\det\pmatrix{6&2\\3&6}=30$. Commented Jan 17, 2021 at 1:38
• Why is $\mathbb{Z}^2/ H = \{(r,s) + H : 0\leq r < 15, 0\leq s < 2\}$? Shouldn't that be $\mathbb{Z}^2/ H = \{(r,s) + H : r,s\in\Bbb Z\}$?
– user870827
Commented Jan 19, 2021 at 21:57

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"): $$\begin{bmatrix} a\\ b \end{bmatrix} \to T \begin{bmatrix} a\\ b \end{bmatrix} = \begin{bmatrix} a\\ b-2a \end{bmatrix} =: \begin{bmatrix} a'\\ b' \end{bmatrix} \to \begin{bmatrix} a'\mod 3\\ b'\mod 10 \end{bmatrix} \in \overset{\displaystyle\Bbb Z/3}{\underset{\displaystyle\Bbb Z/10}\oplus} \cong \Bbb Z/30\ .$$