Efficient presentations

Suppose G is a finite group and rank M(G)=d. In order to be able to obtain an efficient presentation for G one needs to show that G has a presentation on n generators and n+d relations. We describe here a method similar to that used by Campbell and Robertson in [4] which attempts to reduce a given presentation of a group K to an efficient one.

We start with a presentation of K having as small a number of generators and relations as possible. Suppose

K=<X I R>,

X=(xx,...,Xg}, R={rini,...,ri;iit} where ri=ri(xi,...,Xg) and nfs are positive integers. We now try to reduce the number t of relations by considering a stem extension H of K which we hope to be isomorphic to K. To do this, we take two words r^ ^ and rj^j from the set R in such a way that r^ and rj generate the group H defined by

% -H; % n.

H = < X l r ; r . ' , R \ { r ^ ‘, r , ') > .

Then the element r^i^k is in Z(H) because it commutes both with r^ and rj. Now if, in addition, r^^e H' then

r~ e Z(H) n H'

and thus H will be a stem extension of K, that is, a finite group of order at most IM(K)IIKI having K as a homomorphic image. It may happen that IHI=IK1, i.e. H = K. If this is so, we have found a new presentation of K on d generators and t-1 relations. This process of reduction is continued until an efficient presentation for K is obtained.

We note that in constructing a stem extension H of K the condition that r^ and rj generate H is a self-imposed condition to guarantee that the element r^ ^ is central in H. In certain situations, it is possible to combine two arbitrary elements of R and yet obtain a stem extension of K.

Suppose now we have constructed a stem extension H of K. Then the following lemma taken from [4] allows us to determine when H is actually K itself using a coset enumeration program to determine the indices of certain cyclic subgroups of H rather than enumerating the cosets of the trivial subgroup which, in certain cases, exceeds the storage capacity of the machine.

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i

3.5.1 Suppose H is a stem extension of K, he H, and (p the homomorphism from H to K. Let r=l<h(p>l, m=!M(K)l. If (m,r)=l. Then

(i) IHI=rlH:<hm>l;

(ii) if IH:<hP>t=IH:<h>l=IK:<h<p>l for each prime p dividing m, then H = K.

Proof, (i) Let N=Ker(p, s=l<h>l. Then s~rq where q | INI. But INI divides m, by 1.5.13, so q I m. Now l<h"*>l=s/(s,m)=r for (m,r)=I.

(Ü) It follows that l<hP>l=l<h>l=s, for each prime p dividing m. Hence (s,m)=land so q=l. That is, h and hç have the same order. Therefore, H = K.

Let us carry out the foregoing technique for the group S7, the symmetric group. Example 5. Prove that S7 is efficient.

The symmetric group S7 has order 5040 and Schur multiplier the cyclic group of order 2 by 1.5.11 (i). To prove S7 efficient we look for a 2-generator 3-relation presentation. We begin with the following presentation of S7 given in [18] :

K=<X,y I x2=y7=(xy)6=:(xy2xy-2)2=:(xy3xy-3)2=l>. Let Hi be the group obtained by combining two relations as follows:

Hi=<x,y I x2=y7 =(xy^xy-3)2=l, (xy)6=(xy2xy-2)2>

It is easy to check that Hi is generated by xy and xy^xy^ so (xy)b is a central element of Hi. Next we see that H i’=<[x,y],[x,y-l],[x,y2],[x,y3],[x,y3]> and that (xy)6eH i'. Therefore (xy)^ is in Z(Hi)nHi' showing that Hi is a stem extension of K so Hi is either S7 or its covering group by 1.5.13. TC verifies that !Hi:<y>l=720, i.e. H15S7. Now we define

H2=<x,y I x2=y7, (xy5xy^)2=l, (xy)^=(xy2xy2)2>.

A similar verification shows that H2 is a stem extension of Hi ( =87), Using TC we have IH2:<y^>l=720. It follows that H2= S7 by 3.5.1 (i).

Unfortunately, the aim of our reduction method cannot always be acheived. As often happens, on combining the realtions of a given presentation of a group K using the above method we arrive at larger groups -even infinite groups- having K as a

homomorphic image. We therefore need to examine several presentations of K in the hope of finding a suitable one to start with.

Since an exhaustive search for such presentations is not practicable, a step forward might be to try to find a small set of presentations for K to be examined by the reduction method. For 2-generator groups of small orders this could be done using a method suggested by Kenne [31]. In this method a small set of generating paris (x,y) for K are chosen and a presentation on x and y is constructed using Cannon's algorithim [13]. Then those presentaions with fewer relations are taken to be modified.

In chapter 4 where we examine the efficiency of maximal subgroups of simple groups G, IGklQh, we will employ this technique. We note that since we will be dealing with permutation groups of small order, the method will work successfully.

We conclude this section with a CAYLEY program, similar to that in [31], which attemps to find generating pairs for A4XA5

> g :perm(9);

> g.generators: a=(i;2}(3,4h b^(l,2,4), c~(5,7X8,9)4M5,8,6) ; > cl=classes (g);

> FOR i=2 TO length (cl) DO

> FOR TO length (cl) DO

> h=< cl[i], clUJ >;

> IF order (h) EQ order (g)

> THEN print i j , relations (h) ;

> END;

> END; > END;

It is found that class representatives x=(2,4,3)(5,9,7,8,6) and y=(l,2,4)(5,6,7) generate A4XA5 and yield a presentation

<X,y I xl5=y3=(xy)2=(x3y-l)3=l>^

which is an efficient presentation for A4XA5 since M(A4xAg)=C2xC2 by 1.5.12. Note. Suppose H is a subgroup of G generated by two words x=x(a,b), y=y(a,b). By considering the permutation representation of G, we may use a similar program to find a set P of presentations for H. Suppose that Pe P reduces to an efficient presentation for H. Now we require words Wj(a,b), W2(a,b) which generate H and satisfy the presentation P for H. To do so, we first determine the classes of G into which the new generators of H (as permutations) fall ; then PERM can be used to produce two words in