Group presentations - again

My question is about finding presentations for finite groups. It's along similar lines to my earlier question -- but is subtly different! The earlier question is here Group presentations: What's in the kernel of $\phi$?

Let's take a finite group, say $D_4$, and find a set of generators. Say we find two generators; denote them $r$ (geometrically, a rotation through $\pi/2)$ and $s$ (a reflection).

Now let $F$ be the free group on $\{a, b\}$. Consider the homomorphism $\phi$ from $F$ onto $D_4$ with $\phi(a)=r$ and $\phi(b) =s$.

Now we know that $D_4 \cong F/\textrm{ker}(\phi)$. The attached proof (see below) essentially proves that a presentation for $D_4$ is $<a,b \ : \ a^4, b^2, abab>$. This involves showing that $\textrm{ker}(\phi) = N$, where $N$ is the normal subgroup of $F$ generated by the set $\{a^4, b^2, abab\}$.

I notice that the first two relations ($a^4=e$ and $b^2=e$) essentially tell us the order of the generators $a$ and $b$, and the third relation ($abab=e \implies ba=a^3b \implies ab=ba^3$) essentially tells us how to move $a$ and $b$ past each other.

My question is this: can we generalize this idea when finding presentations of a given finite group? If I find a set of generators, I write one relation which represents the order of each generator. I then include one relation for each pair of generators to describe how they move past each other. Will this method always work to produce a presentation of the group? If so, is there a rigorous proof of this?  The answer is no in general. With a finite solvable group (or more generally a polycyclic group, which could be infinite), you can do something like what you say, but you still need a bit more information than just the orders of the generators. I haven't got time to give more details now, but if you google "polycyclic presentation" you should find it. With insolvable groups like $A_5$, I don't think you can do anything like that.
• He said the orders of the generators and one relation for each pair of generators. In a polycyclic presentation, you don't just specify the orders of the generators, the order relations are of the form $x^n=w$, where $w$ is a word in the generators of a smaller subgroup. – Derek Holt Jan 31 '14 at 8:32
Think about this $x^2=y^2=z^2$ and $yx=zy$ and $zx=xz$