Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.