# Presentations: fewer relations than generators implies $G$ infinite.

Take $$r=0$$. Let $$G=\langle a_1\rangle \times ...\times \langle a_n\rangle, \langle a_i\rangle$$ infinite cyclic. $$G$$ is generated by $$n$$ elements and all relations in $$A$$ are relations in $$G$$. Therefor by von Dyck's theorem there is an epimorphism from $$A$$ to $$G$$. But $$G$$ is infinite and so, $$A$$ is infinite too.

For the case $$r>0$$ I have been unable to find a proof. I don't want a solution. I only ask for a hint which allows me to begin working in the problem.

• As a further hint, try defining a nontrivial homomorphism from $A$ to the infinite cyclic group $\langle t \rangle$. So each $x_i$ is mapped to $t^{n_i}$ for some $n_i \in {\mathbb Z}$ and, to apply von Dyck's theorem, the images must satisfy the relations of $A$. So we end up with a system of $r$ linear equations that the $n_i$ must satisfy. – Derek Holt Jan 13 '20 at 8:13
• I define $\lambda: A\to \langle t\rangle$ in the following way: given $g\in A$ write $g$ in normal form, $g=y_1...y_k$ where $y_i\in X=\{x_1,...,x_n\}$. And I write $g\lambda=t^1...t^k$. By the uniqueness of the normal form $\lambda$ is well defined. – stf91 Jan 13 '20 at 17:16
• I MADE A MISTAKE. $g$ is in reduced form, $g=y_1^{\epsilon_1},...,y_k^{\epsilon_k}$. Now, for instance, $(x_2x_5x_2)\lambda=(x_2^2x_5)\lambda=t^4t^5=t^2t^5t^2=(x_2\lambda)(x_5\lambda)(x_2\lambda)$, showing $\lambda$ is a homomorphism. But there is a difficulty. Suppose $x_1x_2=1$ is a relation. Then $t^3=t t^2=(x_1\lambda)(x_2\lambda)= (x_1x_2)lambda=1\lambda=1$ and $t$ has finite order! What am I doing wrong? – stf91 Jan 13 '20 at 17:39
• Suppose I prove the existence of a homomorphism $\alpha:A\to \langle t\rangle$. Then, for instance, if $x_1^2x_2=1$ in $A$, then $t^{2n_1+n_2}=t^{2n_1}t^n_2=(x_1^2x_2)\alpha=1\alpha=1$. But $t$ has infinite order. Therefor $2n_1+n_2=0$. In this way I end up with a system with less equations than unknowns. Such a system, when the $n_i$ belong to field, has infinitely many solutions. I don't know what happens in a ring (the ring $Z$). Anyway, suppose there are infinitely many solutions in $Z$. The set of all these solutions (n-plas in $Z$) constitute a $Z$-module. What do I do now? – stf91 Jan 13 '20 at 20:25
• Once you know that there is a nonzero solution you are done. – Derek Holt Jan 13 '20 at 20:37

## 2 Answers

Hint: $$A$$ is an abelian group, so think of it as a $$\Bbb Z$$-module. $$A$$ can be presented as the quotient of $$\Bbb Z^n$$ by the (free) submodule generated by the $$r$$ relations.

• The presentation given in the problem is an epimorphism from a free group $F$ onto $A$. You mean $F$ is free abelian? – stf91 Jan 12 '20 at 20:33
• Because we already know that $A$ is abelian, we can interpret the setup in the problem as an epimorphism from the free abelian group onto $A$. – Ben Grossmann Jan 12 '20 at 20:35
• Hum... $A=F/R$ where $R$ is the normal closure of the set consisting in the relations. That $F/R$ is abelian does not mean $F$ is abelian. The relations $[x_i,x_j]=1$ are to be interpreted as $[x_i,x_j]\in R$. Why is $F$ abelian? – stf91 Jan 13 '20 at 16:52
• Let $F$ denote the free group on $n$ elements. Define $\tilde F$ to be $F/S$ where $S$ is the normal closure of the relations $[x_i,x_j] = 1$. $A$ can be presented as $\tilde F/R$, where $R$ is the subgroup of $\tilde F$ generated by the $r$ remaining relations – Ben Grossmann Jan 13 '20 at 17:08
• I understand, though I can't see the details. $\bar F$ is a presentation of the free abelian group of rank $n$. So, $\bar F$ itself is free abelian. Let $X=\{x_1,...,x_n\}$ and $s_1(x),...,s_r(x)$ the remaining relations. Then $R=\langle s_1(x)S,...,s_r(x)S\rangle$. And $A$ has the presentation $A=\langle X\mid s_1(x)S,...,s_r(x)S\rangle$. Am I right? – stf91 Jan 13 '20 at 18:34

I get it. I have $$G=\langle m_1,...,m_n\mid s_i(m)\rangle$$ where the $$s_i$$ are the same relations as in $$A$$. For instance, if $$2x_1+x_2=0$$ is a relation in $$A$$ then $$2m_1+m_2=0$$ is a relation in $$G$$. So $$G$$ is a subgroup of $$Z$$ and, as such, it is infinite. Now von Dyck gives an epimorphism $$\beta:A\to G$$ and $$A$$ is infinite.

However, how do I define the homomorphism $$\alpha:A\to \langle t\rangle$$?