Help to find $\frac{\mathbb{Z}\times\mathbb{Z}}{\left<(1,2),(2,3)\right>}$ I can prove that $\frac{\mathbb{Z}\times\mathbb{Z}}{\left<(1,2)\right>}$ is isomorphic to $ \mathbb{Z}$.
Please help me to find $\frac{\mathbb{Z}\times\mathbb{Z}}{\left<(1,2),(2,3)\right>}$
 A: $$\begin{vmatrix}1&2\\2&3\end{vmatrix}=-1\implies \Bbb Z\times\Bbb Z/\langle (1,2)\,,\,(2,3)\rangle\cong 1$$
The above assumes you know something about Smith normal form of matrices, the Fundamental Theorem of finitely generated groups (or of modules over noetherian rings) and stuff.
Another way: show $\;(1,0)\,,\,(0,1)\in\langle (1,2)\,,\,(2,3)\rangle\;$...
A: Define $\phi : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}, (a, b) \mapsto 2a -b.$ Then $\phi$ is a surjective group homomorphism with ker $\phi = (1, 2)\mathbb{Z}.$ (this give you the first isomorphism) Now $\phi (2,3) = 1.$ So using third isomorphism theorem for groups, we have, 
$$\frac{(\mathbb{Z} \times \mathbb{Z})} {\left<(1,2),(2,3)\right>} \cong \frac{\frac{((\mathbb{Z} \times \mathbb{Z})}{\left< (1, 2)\right>}}{\frac{((\left<(1,2),(2,3)\right>)}{\left< (1, 2)\right>}} \cong \frac{\mathbb{Z}}{\mathbb{Z}} = 0.$$
A: Since $$(1,0) = 2 \cdot (1,2) - (2,3)$$
$$(0,1) = -3 \cdot (1,2) + 2 \cdot (2,3)$$
you have that $(1,0),(0,1) \in \langle (1,2), (2,3)\rangle$, so $\mathbb{Z} \times \mathbb{Z} = \langle (1,0), (0,1)\rangle \subseteq \langle (1,2), (2,3)\rangle $.
This means that the quotient is trivial.
