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!

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    $\begingroup$ Consider the $r\times n$ matrix B, where $b_{ij}$ is the power of $x_j$ in the $i$th extra relation. Show there is a nonzero $n\times 1$ column vector of integers $v$ such that $Bv=0$. Perhaps use this to define a map from $A$ to $\mathbb{Z}$? $\endgroup$ – user641 Aug 9 '11 at 6:05
  • $\begingroup$ I don't understand the question. $\endgroup$ – Pierre-Yves Gaillard Aug 9 '11 at 6:33
  • $\begingroup$ @Pierre-Yves Gaillard: I just copied the problem from the book. I don't know how I can explain it clearer. Maybe the notation $[,]$ is not obvious? $[x_i,x_j]$ is the commutator of $x_i$ and $x_j$. $\endgroup$ – ShinyaSakai Aug 10 '11 at 13:34
  • $\begingroup$ @Steve D: Thank you very much. Please allow me to write it more detailed. I hope I am not wrong. ||||| Write $A$ in the addititive form. The $n$ generators are $x_j,j=1, \cdots, n$, and the $r$ relators are $y_i=\sum_{j=1}^nb_{ij}x_j,i=1, \cdots, r$, $b_{ij} \in \mathbb{Z}$. Let the matrix $B=(b_{ij})_{r \times n}$. As $r<n$, the rank of $B$ is less than $n$. There exists a nozero column vector $v=(v_1, \cdots, v_n)^T$ with all the $v_i$'s in $\mathbb{Z}$, such that $Bv=0$. $\endgroup$ – ShinyaSakai Aug 10 '11 at 13:36
  • $\begingroup$ Let $\phi: A \rightarrow \mathbb{Z}$, $\sum_{j=1}^nk_jx_j \mapsto \sum_{j=1}^nk_jv_j$. It is clear that $\phi$ maps the relators to $0$, and is a morphism between the two additive groups. As $v \neq 0$, there is some $v_l \neq 0, 1\leq l \leq n$, then $\phi(A) \supseteq v_l\mathbb{Z}$, so $|A| \geq |v_l\mathbb{Z}|$. $A$ is an infinite group. $\endgroup$ – ShinyaSakai Aug 10 '11 at 13:36

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.

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