Presentations of Finite Simple Groups

Transcription

1 Presentations of Finite Simple Groups Berlin, September / 19

2 Transitive groups SL 2 (p) AGL 1 (p) 2 and 3 (almost) 2 / 19

3 Transitive groups Transitive groups SL 2 (p) AGL 1 (p) One way to obtain a presentations of symmetric/alternating groups with bounded number of is to start with highly symmetric presentation and glue it with a highly transitive group. If we start with the big Coxeter presentation, where the generating set consists of all transpositions, we need to glue with (almost) 4-transitive group in order to obtain a bounded presentations. Unfortunately there are only a few 4-transitive groups which are sufficiently far from Sym(n). It is possible to obtain some bounded presentations using that SL n (F 2 ) acts almost 4 transitively on F n 2 presentations is huge., but the number of in these 3 / 19

4 SL 2 (p) and PSL 2 (p) Transitive groups SL 2 (p) AGL 1 (p) The group PGL 2 (F p ) acts transitively on ordered triples from the set of p + 1 points on the projective line FP 1. Therefore the PSL 2 (p) acts 3-transitively on the p + 1 points if p = 3(mod4), and almost 3-transitively if p = 3(mod4) One can obtain a presentation of SL 2 (p) using that the group SL 2 (Z[1/2]) is finitely presented and does have Congruence Subgroup Property. Thus, adding the relation e p 12 to the presentation of SL 2 (Z[1/2]) yields a presentation of SL 2 (p). Sunday optimized this presentation: PSL 2 (p) = u, t u p = 1, t 2 = (ut) 3, (u 4 tu (p+1)/2 t) 2 = 1, SL 2 (p) has a similar presentation with 2 generators and 2. 4 / 19

5 AGL 1 (p) Transitive groups SL 2 (p) AGL 1 (p) One of the smallest 2-transitive groups is AGL 1 (p) = F p F p acting on F p. It has a presentation with 2 generators and 2 AGL 1 (p) = a, b a p = b p 1, (a s ) b = a s 1 where s(r 1) = 1 for a primitive root r F p. Explanation: The difficult part is to derive the relation a p = b p 1 = 1 There is a similar presentation of the subgroup of index 2 in AGL 1 (p) generated by a and b 2. AGL (2) 1 (p) = a, b a p = b (p 1)/2, (a s ) b = a s 1. 5 / 19

6 for PSL 2 (p) Sym(p + 2) for Sym(p + 2) Presentations for Alt(p + 2) Gluing two copies of Sym(n) or Alt(n) for Sym(n) and Alt(n) Sym(n) and Alt(n) 6 / 19

7 Bounded presentations for PSL 2 (p) Sym(p + 2) for PSL 2 (p) Sym(p + 2) for Sym(p + 2) Presentations for Alt(p + 2) Gluing two copies of Sym(n) or Alt(n) for Sym(n) and Alt(n) The Burnside presentation Sym(n) = s 1, s 2,...,s n 1 s 2 i, (s is j ) 3, (s i s j s i s k ) 2 is invariant under the group Sym(n 1). The group PSL 2 (p) = X R acts almost 3-transitively on the projective line with p + 1 points. Therefore the gluing will give a bounded presentation of PSL 2 (p) Sym(p + 2). If u is an unipotent and h is a generator of the split torus, then X, z R, [z, u], [z, h], z 2, (zz t ) 3, (zz t zz u ) 2, (zz t zz us ) 2 is a presentation of PSL 2 (p) Sym(p + 2), where s is a non-square in F p, (the last relation is redundant if p = 3(mod4)). 7 / 19

8 Bounded presentations for Sym(p + 2) for PSL 2 (p) Sym(p + 2) for Sym(p + 2) Presentations for Alt(p + 2) Gluing two copies of Sym(n) or Alt(n) for Sym(n) and Alt(n) If we add the relation (zu) p+1 to the previous presentation, we obtain a presentation of Sym(p + 2), because PSL 2 (p) Sym(p + 2) PSL 2 (p) Sym(p + 2) and u = u p+1 = (zu) p+1 = 1(mod z G ) This is a bounded presentation, which is NOT short, since few have length p. 8 / 19

9 Presentations for Alt(p + 2) for PSL 2 (p) Sym(p + 2) for Sym(p + 2) Presentations for Alt(p + 2) Gluing two copies of Sym(n) or Alt(n) for Sym(n) and Alt(n) Gluing the CarMichael s presentation with the group AGL 1 (p) we can obtain a presentations of AGL 1 (p) Alt(p + 2) and with 3 generators and 5 : a, b, t a p = b p 1, (a s ) b = a s 1, t 3, t b = t, (tt a ) 2 Again we can kill the AGL 1 (p) factor with one relation, but this requires to passing to a subgroup of index 2 in AGL 1 (p) or using a non-standard embedding of AGL 1 (p) in Alt(p + 2). This might require an extra relation unless p 3(mod4). Tweaking the presentation a little allows us to obtain presentations for Sym(p + 2). If we use PSL 2 (p) instead of AGL 1 (p) we can obtain bounded presentations for Alt(p + 3) and Sym(p + 3). 9 / 19

10 Gluing two copies of Sym(n) or Alt(n) Lemma The symmetric group Sym(m + n k) can be written as for PSL 2 (p) Sym(p + 2) for Sym(p + 2) Presentations for Alt(p + 2) Gluing two copies of Sym(n) or Alt(n) for Sym(n) and Alt(n) Sym(m + n k) = Sym(m) Sym(k) Sym(n)/ [G 1, G 2 ], where G 1 Sym(m) and G 2 Sym(n) are suitably chosen subgroups and k 4. Sketch the proof on the board It is possible to derive all of the form [G 1, G 2 ] using only one relation [x, y] for suitably chosen elements x, y. It is possible to obtain a presentation of Sym(2n k) for 4 < k < n starting from a presentation of Sym(n) and adding 1 generator and / 19

11 Bounded presentations for for PSL 2 (p) Sym(p + 2) for Sym(p + 2) Presentations for Alt(p + 2) Gluing two copies of Sym(n) or Alt(n) for Sym(n) and Alt(n) Using the previous presentations one can obtain presentations for p 3(mod4) of Alt(p + 2) with 3 generators and 6 ; Sym(p + 2) with 3 generators and 6 ; Alt(n) with 4 generators and 10 ; Sym(n) with 4 generators and / 19

12 More efficient gluing 12 / 19

13 More efficient gluing More efficient gluing In the presentation of Alt(p + 2) we can save 1 generator and 2 in the case p = 11(mod12). This is done by using a new generator g such that g 3 = b 2 and g (p 1)/2 = t. Now the [b, t] and t 3 can be dropped. The same works for the presentation of Sym(p + 2) when p = 5(mod6). It is also possible to save another relation by using one relation both for killing the AGL and identifying the intersection in the gluing the two copies of Alt(p + 2). 13 / 19

14 Horner s Rule Short presentations for SL 2 (p) Shortening the Doubling trick Final remarks 14 / 19

15 Horner s Rule Lemma Let u, h G satisfy the relation u h = u 4, then u m can be written as a word on length 4log m on h and u. Horner s Rule Short presentations for SL 2 (p) Shortening the Doubling trick Final remarks Example: u 27 = u = u 42.u 4.2.u 3 = u h2.(u h ) 2.u 3 = (u h u 2 ) h u 3 15 / 19

16 Short presentations for SL 2 (p) Horner s Rule Short presentations for SL 2 (p) Shortening the Doubling trick Final remarks The Horner s rule allows us to modify the Sunday s presentation to a short and bounded one by adding an extra generator h (the diagonal matrix with 2 and 1/2 on the diagonal) and the u h = u 4, h = u 2 (u (p+1)/2 ) t u 2 t and shorten all. It can be shown that any element in SL 2 (p) can be written as a word on length log p in u, h, t. 16 / 19

17 Shortening the Horner s Rule Short presentations for SL 2 (p) Shortening the Doubling trick Final remarks The only long relation left in the presentation of Sym(p + 2) is (zu) p+1 = 1. We can not use the Horner s rule because zu is a p + 1 cycle and can not be conjugated to any of its powers, since p + 1 is not prime. Let k be a generator of the non-split torus in PSL 2 (p) k can be written as a short word on u, t, h and the relation k (p+1)/2 = 1 follows from the other. The element k acts as a product of 2 cycles of lengths (p + 1)/2 on FP 1, thus it can be conjugated to (zu) 2. This conjugation gives an other embedding of PSL 2 (p) into Sym(p + 1), which can be used to shorten the presentation. 17 / 19

18 Doubling trick Horner s Rule Short presentations for SL 2 (p) Shortening the Doubling trick Final remarks Let X R be the short presentation of PSL 2 (p). Thus, inside the group X, X, z R, R, [z, u], [z, h], z 2, (zz t ) 3, (zz t zz u ) 2, z t (zu) 2 z t = k the subgroup generated by z and X is Sym(p + 2), because the last relation implies that zu has order p + 1. Moreover, if it is normal than it is the whole group. Adding the [z, u ] and [z, h ] gives us that the orbits of z under X and X coincide and yields a short and bounded presentation of Sym(p + 2). 18 / 19

19 Final remarks Horner s Rule Short presentations for SL 2 (p) Shortening the Doubling trick Final remarks The previous argument allows us to shorten the presentation of Sym(p + 2). We need to be careful and prove that all involved in the gluing Sym(n) Sym(2n k) can be shortened. One need to use some elementary number theory to show that for every n it is possible to find a prime p = 11(mod12) such that p + 2 < n < 2p. Unfortunately this is not quite true for some n-es less 50 and all these cases have to be done separately. One can not use the doubling trick if we use AGL 1 (p) and the CarMichael presentations. Thus one need to improve the gluing lemma and get alternating groups form symmetric groups. 19 / 19