# An abelian group with $n$ generators and $r(r<n)$ more relations is infinite

Let $$A$$ be an abelian group with generators $$x_1,x_2, \cdots, x_n$$ and defining relations conssisting of $$[x_i,x_j]$$, $$i, and $$r$$ further relations. If $$r, 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!

• 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 $\mathbb{Z}$? – user641 Aug 9 '11 at 6:05
• I don't understand the question. – Pierre-Yves Gaillard Aug 9 '11 at 6:33
• @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 $x_j$. – ShinyaSakai Aug 10 '11 at 13:34
• @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 $Bv=0$. – ShinyaSakai Aug 10 '11 at 13:36
• 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. – 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 $$m_{i1}x_1 + \cdots m_{in}x_n = 0,\quad i=1,\ldots,r Let $$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 $$r), $$\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.