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?
 A: 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.
A: 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.   
