# Prove $(\mathbb{Z} \times \mathbb{Z})/ \langle (2,3)\rangle$ is isomorphic to $\mathbb{Z}$.

I'm trying to prove the following but im stumped:

Prove that $$(\mathbb{Z} \times \mathbb{Z})/\langle (2, 3)\rangle \cong \mathbb{Z}$$.

My attempts so far have been to try and find a single generator of this group. Since its obviously infinite, a single generator would mean its cyclic, and an infinite cyclic group is trivially isomorphic to $$\mathbb{Z}$$ by simply mapping the generator to $$1$$. However, i don't see how its possible for this to have a single generator.

Since $$\gcd(2,3)=1$$ we have $$x,y\in\Bbb Z$$ such that $$2x+3y=1$$. For example, $$x=2,y=-1$$. Now, consider the element $$a:=(y,-x)+\big\langle(2,3)\big\rangle.$$ Now, $$3a=(1,0)+\big\langle(2,3)\big\rangle$$ and $$-2a=(0,1)+\big\langle(2,3)\big\rangle$$. Note that, $$(1,0),(0,1)$$ generates $$\Bbb Z\times\Bbb Z$$. Hence, $$a$$ generates $$\frac{\Bbb Z\times\Bbb Z}{\langle(2,3)\rangle}$$.

Notice that $$3a=(3y,-3x)+\big\langle (2,3)\big\rangle=\big\{(3y,-3x)+(2n,3n)\big|n\in\Bbb Z\big\}$$. So, $$(1,0)-(3y,-3x)=(2x,3x)\in \big\langle(2,3)\big\rangle$$.

• Why does $gcd(2,3)=1$ imply there are $x,y$ s.t. $2x+3y=1$. Why does it even matter? How do you get $3a = (1,0) + <(2,3)>$. I don't understand this proof. Please elaborate further if you can. Jun 21, 2020 at 22:04
• @Hristmar Another way of looking at this proof is to observe that $\{(2,3),(1,2)\}$ form a basis of $\Bbb{Z}\times\Bbb{Z}$. Surely you agree that if you mod out $\langle(1,0)\rangle$ you get $\Bbb{Z}$ as the quotient. It's the same thing here. Jun 22, 2020 at 10:53

Let's look for a surjective homomorphism $$\varphi\colon \mathbb{Z}\times\mathbb{Z}\to \mathbb{Z}$$ such that $$\operatorname{ker}\varphi=\langle (2,3)\rangle$$. If we succeed, then by the First Homomorphism Theorem we are done.

Let's define:

$$\varphi(m,n):=3m-2n \tag 1$$

Then:

\begin{alignat}{1} \varphi((m,n)+(k,l)) &= \varphi(m+k,n+l) \\ &=3(m+k)-2(n+l) \\ &=(3m-2n)+(3k-2l) \\ &=\varphi(m,n)+\varphi(k,l) \\ \tag 2 \end{alignat}

and $$\varphi$$ is a homomorphism.

Furthermore:

\begin{alignat}{1} &\forall r\in \mathbb{Z}, \space \exists (m,n)\in \mathbb{Z}\times\mathbb{Z}\mid r=3m-2n \iff \operatorname{gcd}(3,2)=1 \tag 3 \end{alignat}

which is the case, whence $$\varphi$$ is surjective.

Finally:

\begin{alignat}{1} \operatorname{ker}\varphi &= \{(m,n)\in \mathbb{Z}\times\mathbb{Z}\mid\varphi(m,n)=0\} \\ &= \{(m,n)\in \mathbb{Z}\times\mathbb{Z}\mid 3m-2n=0\} \\ &= \biggl\{(m,n)\in \mathbb{Z}\times\mathbb{Z}\mid n=\frac{3}{2}m\biggr\} \\ &= \biggl\{\biggl(m,\frac{3}{2}m\biggr)\mid m\in 2\mathbb{Z}\biggr\} \\ &= \{(2r,3r)\mid r\in\mathbb{Z}\} \\ &=\langle (2,3) \rangle \\ \tag 4 \end{alignat}

Let $$I = \langle (2,3) \rangle = \{\cdots, (0,0), (2,3), (4,6), (6,9), \cdots\}$$

Then $$R = (\mathbb Z \times \mathbb Z) / I$$ is the ring of elements $$\{ (x,y) + I \,|\, (x,y) \in \mathbb Z \}$$.

So, as elements of $$R$$:

• $$(0,0) = (2,3) = (4,6) = (6,9)$$
• $$(1,0) = (3,3) = (5,6) = (7,9)$$
• $$(2,0) = (4,3) = (6,6) = (8,9)$$
• $$(3,0) = (5,3) = (7,6) = (9,9)$$
• $$(0,1) = (2,4) = (4,7) = (6,10)$$ and going down $$= (-2,-2)$$
• $$(0,2) = (2,5) = (4,8) = (6,11)$$ going down $$= (-2,-1) = (-4, -4)$$
• $$(0,3) = (2,6) = (4,9) = (6,12)$$
• $$(1,2) = (3,5) = (5,8) = (7,11)$$
• $$(1,3) = (3,6) = (5,9) = (7,12)$$

It seems like every element can be represented by an element $$(r,r)$$

Given $$(x,y)$$ you can reduce it to a unique representative $$(r,r)$$ by adding or subtracting a multiple of $$(2,3)$$. So $$(r,r) = (x,y) + k\cdot(2,3)$$ gives us the system

• $$r = x + 2 k$$
• $$r = y + 3 k$$

which we can solve

$$k = \frac{x - y}{3 - 2} = x - y$$