# Finitely presented group with fewer relations than generators.

I some how feel that, any finitely presented group with less relations than generators has to be an infinite group.

In one of my questions, in an answer I have seen a group which is finitely presented but it is not finite.

$$G=\langle a,b\mid a^2=b^2=1\rangle$$ this group is infinite. I see that if there is at least one more relation in between $$a$$ & $$b$$ then the group is finite.

I claim that if there are fewer relations than generators of a finitely presented group, then the group has to be infinite.

I would like to see is it true in general??

Is any finitely presented group with fewer relations than generators an infinite Group?

Please suggest some idea to see if this is true/false?

• Just making sure: Are you counting $a^2=b^2=1$ as two relations? Commented Aug 29, 2013 at 7:10
• @ChrisCulter No No :P That is only one relation :P
– user87543
Commented Aug 29, 2013 at 7:12
• Ah, well, I think it's standard to count that as two relations. One way to make this clearer would be to write every relation as setting something equal to the identity, so $G=\langle a,b\mid a^2,b^2\rangle$. Under what criteria could you count it as only one relation? Commented Aug 29, 2013 at 7:18
• @PraphullaKoushik: The relation $b^{-2}a^2=1$ is not equivalent to the two relations $a^2=1$ and $b^2=1$. Commented Aug 29, 2013 at 7:23
• Rule of thumb: The number of relations is equal to the number of equality signs, so $a^2=b^2=1$ is two relations :-) Commented Aug 29, 2013 at 7:51

In the answer I will use the standard way to number the relators in a presentation. Then, if $$G$$ is a group of deficiency $$\ge 1$$ (i.e., admits a presentation with $$n$$ generators and $$k$$ relators, where $$n>k$$) then $$G$$ is infinite and, moreover, admits an epimorphism to the infinite cyclic group. To prove this, consider the rational vector space $$V=Hom(G, {\mathbb Q})$$ (where we regard $${\mathbb Q}$$ as the additive group of the field $${\mathbb Q}$$):

This vector space $$V$$ is given by imposing $$k$$ linear equations on $$Hom(F_n, {\mathbb Q})={\mathbb Q}^n$$, since every homomorphism to $${\mathbb Q}$$ is determined by its values on generators of $$G$$, while the only restrictions on the images of generators are that each relator maps to zero: Every such condition is one linear equation.

Hence, $$dim Hom(G, {\mathbb Q}) \ge n-k\ge 1$$. It therefore, follows that there exists a nonzero homomorphism $$h: G \to {\mathbb Q}$$. The image of this homomorphism is an infinite torsion free finitely generated subgroup (as $${\mathbb Q}$$ contains no nonzero finite subgroups), i.e. $${\mathbb Z}^r$$ for some $$r\ge 1$$. Since it is a subgroup of $${\mathbb Q}$$, $$r=1$$. Thus, $$G$$ admits an epimorphism $$h: G\to {\mathbb Z}$$, and, hence, is an infinite group. In particular, $$G$$ contains an element of infinite order (any element $$g\in G$$ such that $$h(g)\ne 0$$).

In fact, one can prove more, namely that abelianization of $$G$$ has rank $$\ge n-k$$, but we do not need this.

A much more interesting result is due to Baumslag and Pride: They proved that every group of deficiency $$\ge 2$$ has a finite index subgroup which admits an epimorphism to a nonabelan free group. Such a group is called large. See also http://arxiv.org/pdf/1007.1489.pdf and references there.

• I am not familiar with "abelianization of a group" Could you edit so that to make it less complicated :)
– user87543
Commented Aug 29, 2013 at 8:55
• @PraphullaKoushik: Done. Commented Aug 29, 2013 at 9:36
• somehow everyone is trying to explain it in detail. but, I am unable to grasp it. May be it is too high for me. No Idea :(
– user87543
Commented Aug 29, 2013 at 11:00
• @PraphullaKoushik: Maybe you should just study group theory a bit more; you can start by reading Lyndon and Schupp "Combinatorial Group Theory" (I found it a very nice introduction to the subject). Commented Aug 29, 2013 at 13:28
• Indeed, answers here do not replace the onus on OPs to actually pick up a book and gather the standard lore of the subject! Commented Aug 29, 2013 at 16:10

I think you are somehow looking for Deficiency of a group and here. I had a course before about the presentation of groups and in my notes it's stated a Theorem as follows. I couldn't find it through web. Sorry.

Theorem (Mc Donald): Every group $G$ with positive deficiency is infinite.

You may consult here for more examples. Of course, the converse of the above theorem is not right in general. See here and here for finding the counter-examples.