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

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.
