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Let $X=\{x_1,...,x_n\}$ be the set of generators and $\Delta=\{s_1(x),...,s_r(x)\}$ be the set of relations. There is a homomorphism $G\to \langle t\rangle$, where $\langle t \rangle$ is the infinite cyclic group, that sends every generator into $t$. It is easy to prove. So, the set of all the nontrivial homomorphisms from $G$ to $\langle t\rangle$ is not empty. Let $\theta$ be one such homomorphism. Then $\theta$ sends $x_i$ into $t^{n_i}$ for some $n_i\in Z$. Now suppose $x_1^2x_3^{-1}=1$ is a relation in $G$. Then

$$\begin{align}t^{2n_1-n_3}&=t^{2n_1}t^{-n_3}\\ &=(x_1^2)\theta (x_3^{-1})\theta\\ &=(x_1^2x_3^{-1})\theta\\ &=1\theta\\ &=1. \end{align}$$

But $t$ has infinite order. So $2n_1-n_3=0$. More generally in this way I get a system of $r$ equations in $n$ unknowns with $r<n$. I know this system has a solution in the reals. I'll assume it has a solution in the integers. I'm not sure. If it has, then I have $L=\langle m_1,...,m_n: s_1(m),...,s_r(m)\rangle$. Here $L$ is generated by $n$ elements which satisfy the same relations as in $G$ and, by von Dyck's theorem, there is an epimorphism $\varphi:G\to L$. But $L$ is a subgroup of $Z$ the integers and, as such, it is infinite. Therefor $G$ is infinite.

There remains one thing to be proved. That a system of $r$ linear equations in $n$ unknowns and coefficients in $Z$ with $r<n$ has a solution in the integers. Honestly I do not know if this is even true. Assuming it is true, is the proof valid?

EDIT: I think the system has a solution in the rationals. Then I multiply each equation by the least common multiple of the denominators and I get a solution in the integers.

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    $\begingroup$ You seem to have answered your own question. Yes, the proof is valid. $\endgroup$ – Derek Holt Jan 14 '20 at 22:21
  • $\begingroup$ See also this post. $\endgroup$ – Dietrich Burde Jan 14 '20 at 22:24
  • $\begingroup$ I don't understand why there is a non-trivial map to $\mathbb{Z}$. If one of the relations is $x_1^2 x_3^{-1} = 1$, then the map which sends each $x_i$ to $t$ is not well defined. $\endgroup$ – Jason DeVito Jan 14 '20 at 22:44
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    $\begingroup$ Well, if $G$ is a group with $n$ generators and $r < n$ relations, then $G^{\mathrm{ab}}$ is an abelian group with $n$ generators and $r < n$ relations. So if the argument there shows that $G^{\mathrm{ab}}$ is infinite, then $G$ also certainly has to be infinite. $\endgroup$ – Daniel Schepler Jan 15 '20 at 3:17
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    $\begingroup$ Does this answer your question? Finitely presented Group with less relations than Generators. $\endgroup$ – Moishe Kohan Jan 15 '20 at 3:27
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Your proof has good ideas, but it's executed in a somewhat sketchy and confusing way. To put your argument simply, you could define the "degree" of a generator $x_i$ in a relation $w=1$ to be the signed number of times $x_i$ appears (i.e. the value of $w$ under the homomorphism taking $x_i$ to $1\in\mathbb Z$ and taking every other generator to $0$) - call this $d_{w,i}$. If you could find a sequence $n_i$ such that for every relation $d_{w,i}$ we had $$\sum_in_i\cdot d_{w,i} = 0$$ it would be true that there was a homomorphism $f:G\rightarrow\mathbb Z$ such that $$f(x_i)=n_i$$ due to the universal property (a.k.a. von Dyck's theorem) that the group $\langle x_1,\ldots, x_n | w_1,\ldots, w_r\rangle$ has, for group $G'$ and every assignment of values $\bar x_i\in G$ satisfying the relations $w_i$ in $G'$, a unique map $f:G\rightarrow G'$ such that $f(x_i)=\bar x_i$. So long as $f$ is not the zero map, its image is not trivial, hence must be infinite as the image is a subgroup of $\mathbb Z$.

You can establish the existence of such a assignment of values $n_i$ by linear algebra: first, there is such a solution in the rational numbers because there are $r$ linear relations in a space of dimension $n$, hence are satisfied in some subspace of dimension $n-r > 0$ - in particular, must have a non-trivial rational solution. However, you can always multiply out the denominators of a rational solution to get an integer solution.

You can also make this argument by considering the abelianization of $G$ - which is then $\mathbb Z^d$ modulo the relations $\sum_i x_i\cdot d_{w,i} = 0$. The linear algebra argument then applies a little more directly, since then the abelianization of $G$ is $\mathbb Z^d$ modulo some subgroup generated by $r$ terms, and then you can just apply the prior paragraph's argument.

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Indeed, the supposed existence of a nontrivial homomorphism $G \to \mathbb{Z}$ suffices. Since subgroups of $\mathbb{Z}$ are infinite cyclic, we may suppose the map is surjective. Free groups are projective objects in the category of groups, thus the map $G \twoheadrightarrow \mathbb{Z}$ admits a section. $G$ thus contains an infinite cyclic subgroup and is itself infinite.

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    $\begingroup$ Even more simply, a group cannot map onto a group with infinitely many elements unless it itself is infinite. $\endgroup$ – Rylee Lyman Jan 15 '20 at 3:25

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