# Suppose that $G$ is a group with $n$ generators and $r$ relations whether $r < n$. Prove that $G$ is infinite. [duplicate]

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 $$r. I know this system has a solution in the reals. I'll assume it has a solution in the integers. I'm not sure. If it has, then I have $$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 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.

• You seem to have answered your own question. Yes, the proof is valid. – Derek Holt Jan 14 '20 at 22:21
• See also this post. – Dietrich Burde Jan 14 '20 at 22:24
• 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. – Jason DeVito Jan 14 '20 at 22:44
• 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. – Daniel Schepler Jan 15 '20 at 3:17
• Does this answer your question? Finitely presented Group with less relations than Generators. – Moishe Kohan Jan 15 '20 at 3:27

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.
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.