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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.

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3 Answers 3

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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, $(1,0)-(3y,-3x)=(2x,3x)\in \big\langle(2,3)\big\rangle$.

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  • $\begingroup$ 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. $\endgroup$
    – user569685
    Jun 21, 2020 at 22:04
  • $\begingroup$ @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. $\endgroup$ Jun 22, 2020 at 10:53
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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}

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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

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

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