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Let $G$ be finitely presented group with $n$ generators and $r$ relations where $n>r$. I want to show that $G$ has an element with infinite order.

My attempt:

Assume that $F_n$ is free group with $n$ generators. We know that $F_n/F^{\prime}_n\cong\mathbb Z ^n$. So $F_n$ has an element of infinite order, but I don't know how use the relations?

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  • A homogeneous system of linear equations with more unknowns than equations has a nonzero solution. Use this to show that $G$ has a nontrivial homomorphism onto ${\mathbb Z}$ and deduce the result. – Derek Holt Aug 11 '14 at 17:04
  • @DerekHolt, how? – user108209 Aug 11 '14 at 17:29
  • Let $a_1,\ldots,a_n$ be the group generators. Then $a_i \mapsto x_i$ (with $x_i \in \mathbb{Z}$) extends to a homomorphism $G \to {\mathbb Z}$ if and only if the $x_i$ satisfy the $r$ group relations, or equivalently if the $x_i$ satisfy $r$ homogeneous linear equations. Since $n>r$, these equations have a nonzero solution in ${\mathbb Q}$ and hence also in ${\mathbb Z}$, so we get a nontrivial homomorphism $G \to {\mathbb Z}$. – Derek Holt Aug 11 '14 at 20:39
  • @DerekHolt, thanks for solution. Does a finitely presented group exist such that it has an element with infinite order for case n<r? – user108209 Aug 12 '14 at 06:12
  • @user108209, one way of doing it is by making sure that one of the generators does not occur in any of the relations. – Andreas Caranti Aug 12 '14 at 15:49
  • https://math.stackexchange.com/questions/478841/finitely-presented-group-with-fewer-relations-than-generators?rq=1 – Moishe Kohan Jan 22 '20 at 22:44

1 Answers1

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Hint: Consider the abelianisation of your group. This is a finitely generated abelian group, and you want to prove that its decomposition contains a $\mathbb{Z}$-term. Consider your abelian group as a $\mathbb{Z}$-module, form an appropriate matrix and diagonalise. What do you notice?

Sorry, I do not got the time to expand on this more just now. But my point is basically: compute the abelinisation, and do what you see here. What do you notice?

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