TITLE: surjective group homomorphism between $\mathbb{Z}^2$ and $\mathbb{Z}_{30}$
QUESTION [5 upvotes]: 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.

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