An abelian group with $n$ generators and $r(r
Let $A$ be an abelian group with generators $x_1,x_2, \cdots, x_n$ and defining relations conssisting of $[x_i,x_j]$, $i<j=1,2, \cdots, n$, and $r$ further relations. If $r<n$, prove that $A$ is infinite.
Let $F$ be the free abelian group on $n$ generators. Do I have to prove the $r$ relations generate a finite subgroup of $F$? How to?
Thank you very much!
 A: You can use linear algebra to explicitly find an element of $A$ with infinite order:
Suppose $A$ is generated by $x_1,\ldots,x_n$ subject to relations
$$
m_{i1}x_1 + \cdots m_{in}x_n = 0,\quad i=1,\ldots,r<n.
$$
Let $M=(m_{ij})\in R^{r\times n}$.
Let $F$ be the free abelian group on the $n$ generators $x_1,\ldots,x_n$. Then there is a unique surjection $\phi:F\to Z^n/C_Z(M^T)$ such that $\varphi(x_i)=e_i$ for all $i=1,\ldots,n)$, where $e_i$ is the $i$-th standard basis vector of $Z^n$ and $C_Z(M^T)$ denotes the integral column space of $M^T$, i.e., the set of all $Z$-linear combinations of columns of $M^T$. By design, $\phi$ vanishes on the subgroup $G$ of $F$ generated by the set $$\{m_{i1}x_1+\cdots + m_{in}x_n:i=1\ldots,r\}.$$ Therefore, $\varphi$ descends to a homomorphism $$\bar{\varphi}:A=F/G\to Z^n/C_Z(M^T).$$
Since $\varphi$ is surjective, so is $\bar{\varphi}$. Thus, it suffices to show that $Z^n/C_Z(M^T)$ is infinite.
Since $M$ has fewer rows than columns (here we use the inequality $r<n$),
$$
\operatorname{dim}N(M)> 0\quad(\text{dimension over }\mathbb{Q}).
$$
Here, $N(M)$ means nullspace over $\mathbb{Q}$.
Let $x\in N(M)$ be nonzero and let $k$ be a integer such that $y :=kx\in Z^n$. Therefore, $Zy$ is an infinite subset of $N(M)\cap Z^n$.
By a fundamental theorem of linear algebra, $N(M)$ is the orthogonal complement of $C(M^T)$ (column space over $\mathbb{Q}$). In particular, $N(M)\cap C(M^T)=\{0\}$. It follows that
$$Zy \cap C_Z(M^T) \subseteq Zy\cap C(M^T) = \{0\}.$$
This implies that the restriction $$\phi|_{Zy}:Zy\to Z^n/C_Z(M^T)$$ is injective.
Since $Zy$ is infinite, $Z^n/C_Z(M^T)$ must be infinite, too, as was to be shown.
Remark: By taking a basis of $N(M)$ (row reduction!) and clearing denominators, rather than just a single nonzero vector, you can conclude that $A$ has rank at least $n-\operatorname{rank}(A)$ by producing an explicit subgroup of that rank.
