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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.

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    $\begingroup$ 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. $\endgroup$ – Derek Holt Jan 13 '20 at 8:13
  • $\begingroup$ 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. $\endgroup$ – stf91 Jan 13 '20 at 17:16
  • $\begingroup$ 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? $\endgroup$ – stf91 Jan 13 '20 at 17:39
  • $\begingroup$ 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? $\endgroup$ – stf91 Jan 13 '20 at 20:25
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    $\begingroup$ Once you know that there is a nonzero solution you are done. $\endgroup$ – Derek Holt Jan 13 '20 at 20:37
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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.

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  • $\begingroup$ The presentation given in the problem is an epimorphism from a free group $F$ onto $A$. You mean $F$ is free abelian? $\endgroup$ – stf91 Jan 12 '20 at 20:33
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    $\begingroup$ 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 $A$. $\endgroup$ – Ben Grossmann Jan 12 '20 at 20:35
  • $\begingroup$ 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? $\endgroup$ – stf91 Jan 13 '20 at 16:52
  • $\begingroup$ 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 $\endgroup$ – Ben Grossmann Jan 13 '20 at 17:08
  • $\begingroup$ 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? $\endgroup$ – stf91 Jan 13 '20 at 18:34
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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 $\alpha:A\to \langle t\rangle$?

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