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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

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?

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2 Answers 2

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.

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The OP is talking about more than just the orders of the generators. [And just writing the orders wouldn't suffice to characterize any finite group except cyclic ones anyway; you would end up with free products of cyclic groups, which are infinite.] –  anon Jan 30 at 22:50
    
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 at 8:32
    
Thanks Derek. Do you have a recommendation for a book which gives a good introductory grounding in group presentations? I'm only at the second year undergraduate level, so I'm not looking for something very technical or specialised. Maybe just a general group theory book but one which treats groups presentations quite thoroughly. –  user123473 Feb 1 at 0:26
    
The books by D.L. Johnson on group presentations are quite good, but might still be difficult at the second year undergraduate level. Although it is an important topic, it is difficult to treat it thoroughly at an elementary level, because you need first to define free groups and prove some of their basic properties. –  Derek Holt Feb 1 at 18:28
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Trivially for a finite group you can have the generators be the entire group and the relations on pairs of generators would just be the multiplication table.

And that of course defines the group.

Think about this $x^2=y^2=z^2$ and $yx=zy$ and $zx=xz$

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