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Consider the group $\DeclareMathOperator{\bZ}{\mathbb{Z}}\DeclareMathOperator{\bN}{\mathbb{N}}G=\bZ\times\bZ/H$ where $H=\langle(3,-2)\rangle$ (in general it looks like it works the same for subgroups $(m,n)$ where $m,n$ are coprime).

I want to know about the structure of $G$. If I am not mistaken, $G$ is cyclic: If $x=(1,0)+H,y=(0,1)+H$ then $2(-x+y)=-2x+2y=x$ and $3(-x+y)=x-x+y=y$.

Also $G$ is torsion-free, i.e., for any $n\in\bZ$, $$n(-x+y)\neq 0$$ since for small $n$ you get $x$ and $y$ as above, and for larger $|n|$ you get greater (positive or negative) coefficients in $x$ when letting the coefficients of $y$ range from $-1$ to $1$ (so it cannot be $0$).

So it seems to me that $G\cong \bZ$. Am I mistaken?

I'm conflicted, because I'm under the impression that $G$ is the Grothendieck group of the monoid $M=\bN\times\bN/\sim$ where $(3,0)\sim (0,2)$.

So... is that wrong? Because $M$ is not torsion-free and $G$ is. Or am I missing something else?

Edit: About torsion in monoids, we say $M$ is torsion-free if for any $n\geq 1$, and $m,m'\in M$, $nm=nm'\Rightarrow m=m'$. In groups this implies the usual definition of being torsion-free. This way we can say that $M$ is torsion-free iff its Grothendieck group $G$ is torsion-free: $nm=nm'$ in $M$ means that $n(m-m)'=0$ in $G$ and viceversa.

Edit 2: Looks like I'm mistaken in that $M$ is not torsion free.

LeviathanTheEsper
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    I'm probably being stupid, but what torsion elements are there in $M$? – Rob Arthan Oct 02 '21 at 22:40
  • Since the group has two generators and one relation, it will always be infinite. Since no element of $\mathbb Z\times \mathbb Z$ powers to a multiple of $(3,-2)$ that isn't already a multiple of $(3,-2)$, there cannot be an element of finite order in $G$. Thus $G\cong \mathbb Z$. – David A. Craven Oct 02 '21 at 22:40
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    @DavidA.Craven Just because it's torsion free doesn't mean it's $\mathbb{Z}$ it could be $\mathbb{Z}\times\mathbb{Z}$ – Mars Oct 02 '21 at 22:47
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    @Mars No. finitely generated abelian groups are never isomorphic to proper quotients of themselves. (They are Hopfian.) – David A. Craven Oct 02 '21 at 22:50
  • @RobArthan Torsion in monoids is a little more general. We say $M$ is torsion-free if for any $n\geq 1$, and $m,m'\in M$, $nm=nm'\Rightarrow m=m'$. In groups this implies the usual definition of being torsion-free. See the edit for more info. – LeviathanTheEsper Oct 03 '21 at 00:22
  • The general procedure for doing this uses the Smith normal form of the relations matrix. Here are some related posts: 1, 2 – Viktor Vaughn Oct 03 '21 at 00:47
  • I think you mean $n \neq 0$ not $n \ge 1$. I would call what you call a torsion-free monoid a cancellative monoid. But my question still stands: why isn't your $M$ torsion-free (or cancellative). – Rob Arthan Oct 03 '21 at 00:53
  • @RobArthan I meant $n\geq 1$, because you cannot necessarily define $-m$ for $m\in M$. Also, cancellative monoid is already used for the property $n+m=n+m'\Rightarrow m=m'$.

    But yes, it looks like I mistook and $M$ is indeed torsion-free despite of what I thought before.

     – LeviathanTheEsper Oct 03 '21 at 00:56
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    I thought you were using multiplicative notation for your monoid and that $n$ was in $M$ not $\Bbb{N}$. – Rob Arthan Oct 03 '21 at 01:02
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    I am pretty sure that the general question is already solved on M.SE, that is, $\mathbb Z\times\mathbb Z/\left<(a,b)\right>\simeq\mathbb Z\times\mathbb Z_d$ where $d=\gcd(a,b)$. – user26857 Oct 03 '21 at 15:54

2 Answers2

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Because three and two are coprime, there is a matrix in $\mathrm{SL}(2,\mathbb Z)$ with first row your vector. This means here is an automorphism of $\mathbb Z^2$ that sends $(3,-2)$ to one of the canonical basis vectors, so your group is $\mathbb Z$. This extends of course to taking any quotient of a power of $\mathbb Z$ by a row of a matrix in the special linear group.

Add: it is good to point out that in this case you can also use the algorithm to reach the Smith Normal Form on the matrix $(3, -2)$, which gives the following:

$$(3, -2)\to (1, -2) \to (1, 0)$$

as it will do for any pair of coprime integers. More generally, the SNF of $(a,b)$ will be $(d, 0)$ where $d=\mathrm{gcd}(a,b)$ and the quotient will be isomorphic to $\mathbb Z\oplus\mathbb Z/d$.

user26857
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Pedro
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You're correct.

Sanity check:

gap> F:=FreeGroup(2);
<free group on the generators [ f1, f2 ]>
gap> gens:=[F.1*F.2*F.1^(-1)*F.2^(-1), F.1^3F.2^(-2)];
[ f1*f2*f1^-1*f.2^-1, f1^3*f2^(-2) ]
gap> G:=F/gens;
<fp group on the generators [ f1, f2 ]>
gap> StructureDescription(G);
"Z"
gap>
Shaun
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