Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Finite simple groups (non-abelian) can generated by two elements.

Let $G=\langle x,y|x^l=y^m=(xy)^n=1,...\rangle$ be a finite simple group (non-abelian), and $\langle x,y|x^p=y^q=(xy)^r=1,...\rangle$ be another presentation of $G$. (Here, "..." means possibly more relations).

1) If $(1/l)+(1/m)+(1/n)<1$, then does it imply that $(1/p)+(1/q)+(1/r)<1$?

2) If $(1/l)+(1/m)+(1/n)=1$, then does it imply that $(1/p)+(1/q)+(1/r)=1$?

3) If $(1/l)+(1/m)+(1/n)>1$, then does it imply that $(1/p)+(1/q)+(1/r)>1$?

(For a group $G=\langle x,y|x^l=y^m=(xy)^n=1,...\rangle$ to be a group of symmetries of a compact orientable surface, there are some restrictions on the integers $l,m,n$ and genus of surface, due to Riemann-Hurwitz relation. If the integers (l,m,n) satisfy some conditions as above, then it will allow us to consider less number of presentations of finite simple groups to check for possibility of action of the group on the surface).

share|improve this question

3 Answers 3

up vote 7 down vote accepted

No. The simple group of order 60 is a counterexample to (1) and (3).

If {x,y} is a generating set of G, call its signature 1/|x| + 1/|y| + 1/|xy|. It appears that the signature of most generating sets of non-abelian simple groups are less than 1.

I'll assume that l,m,n are required to be the exact orders (and are not just sneakily enlarged to make it easier to get < 1, since it is already easy to get < 1). Hence you are just asking for some sort of "law of inertia" for the "signature" of two-generator sets.

However, the simple group of order 60 has the following signatures: 3/5, 11/15, 13/15, 9/10, 31/30.

It does however appear to be a bit rare. The simple groups I've checked other than A5 all have all signatures less than 1.

I didn't see any simple groups satisfying the hypothesis of (2), so I guess it could be vacuously true.


Maybe this table of signatures up to order 10000 will help:

  • 60 PSL(2,4): 3/5, 11/15, 13/15, 9/10, 31/30
  • 168 PSL(2,7): 3/7, 15/28, 13/21, 9/14, 61/84, 3/4, 11/14, 17/21, 5/6, 25/28, 11/12, 41/42
  • 360 PSL(2,9): 3/5, 13/20, 7/10, 11/15, 3/4, 47/60, 13/15, 9/10, 11/12, 19/20
  • 504 PSL(2,8): 1/3, 23/63, 25/63, 3/7, 5/9, 37/63, 13/21, 13/18, 95/126, 7/9, 11/14, 17/21, 17/18, 41/42
  • 660 PSL(2,11): 3/11, 23/66, 21/55, 14/33, 151/330, 27/55, 1/2, 17/33, 8/15, 17/30, 13/22, 3/5, 103/165, 2/3, 15/22, 7/10, 11/15, 25/33, 87/110, 5/6, 13/15, 9/10, 61/66
  • 1092 PSL(2,13): 3/13, 27/91, 25/78, 33/91, 211/546, 16/39, 3/7, 19/42, 10/21, 19/39, 1/2, 151/273, 15/26, 13/21, 9/14, 17/26, 2/3, 131/182, 29/39, 11/14, 17/21, 5/6, 71/78, 41/42
  • 2448 PSL(2,17): 3/17, 35/153, 33/136, 43/153, 361/1224, 21/68, 1/3, 25/72, 13/36, 25/68, 3/8, 257/612, 59/136, 23/51, 17/36, 35/72, 1/2, 77/153, 211/408, 5/9, 19/34, 41/72, 7/12, 11/18, 21/34, 5/8, 131/204, 205/306, 93/136, 25/36, 17/24, 13/18, 37/51, 53/72, 3/4, 7/9, 19/24, 55/68, 5/6, 31/36, 7/8, 91/102, 11/12, 17/18, 23/24
  • 2520 Alt(7): 3/7, 19/42, 10/21, 17/35, 1/2, 107/210, 8/15, 15/28, 19/35, 47/84, 17/30, 7/12, 83/140, 3/5, 37/60, 13/21, 9/14, 13/20, 2/3, 71/105, 7/10, 61/84, 11/15, 3/4, 47/60, 11/14, 17/21, 5/6, 59/70, 13/15, 25/28
  • 3420 PSL(2,19): 3/19, 39/190, 37/171, 24/95, 451/1710, 47/171, 3/10, 29/95, 14/45, 29/90, 1/3, 67/190, 311/855, 2/5, 37/90, 19/45, 25/57, 43/95, 277/570, 85/171, 1/2, 23/45, 8/15, 49/90, 5/9, 167/285, 3/5, 23/38, 19/30, 29/45, 62/95, 227/342, 7/10, 32/45, 41/57, 13/18, 11/15, 143/190, 23/30, 7/9, 4/5, 73/90, 13/15, 101/114, 9/10, 14/15, 17/18
  • 4080 PSL(2,16): 3/17, 47/255, 49/255, 1/5, 27/85, 83/255, 1/3, 23/51, 39/85, 7/15, 151/255, 3/5, 21/34, 319/510, 19/30, 37/51, 11/15, 129/170, 23/30, 91/102, 9/10
  • 5616 PSL(3,3): 3/13, 29/104, 25/78, 17/52, 115/312, 3/8, 21/52, 16/39, 5/12, 47/104, 11/24, 19/39, 77/156, 1/2, 167/312, 13/24, 15/26, 7/12, 5/8, 17/26, 103/156, 2/3, 73/104, 17/24, 29/39, 3/4, 19/24, 43/52, 5/6, 71/78, 11/12
  • 6048 PSU(3,3): 1/4, 7/24, 13/42, 1/3, 59/168, 31/84, 3/8, 11/28, 23/56, 5/12, 3/7, 73/168, 19/42, 11/24, 10/21, 1/2, 29/56, 15/28, 13/24, 47/84, 7/12, 101/168, 13/21, 5/8, 9/14, 2/3, 17/24, 61/84, 3/4, 43/56, 11/14, 19/24, 17/21, 5/6
  • 6072 PSL(2,23): 3/23, 47/276, 45/253, 29/138, 661/3036, 57/253, 1/4, 35/138, 17/66, 35/132, 3/11, 27/92, 457/1518, 1/3, 31/92, 15/44, 23/66, 26/69, 389/1012, 5/12, 29/69, 14/33, 19/44, 127/276, 355/759, 1/2, 67/132, 17/33, 25/46, 7/12, 27/46, 13/22, 173/276, 321/506, 2/3, 89/132, 15/22, 49/69, 3/4, 25/33, 73/92, 5/6, 37/44, 121/138, 11/12, 61/66
  • 7800 PSL(2,25): 3/13, 37/156, 19/78, 1/4, 25/78, 17/52, 1/3, 23/65, 281/780, 11/30, 21/52, 16/39, 5/12, 173/390, 9/20, 31/65, 29/60, 19/39, 77/156, 1/2, 137/260, 8/15, 17/30, 15/26, 7/12, 119/195, 37/60, 13/20, 17/26, 103/156, 2/3, 7/10, 29/39, 3/4, 101/130, 47/60, 43/52, 5/6, 71/78, 11/12
  • 7920 M11: 3/11, 27/88, 15/44, 23/66, 3/8, 21/55, 101/264, 183/440, 5/12, 14/33, 19/44, 9/20, 151/330, 11/24, 41/88, 27/55, 59/120, 1/2, 67/132, 17/33, 21/40, 8/15, 119/220, 13/24, 145/264, 17/30, 23/40, 7/12, 13/22, 3/5, 37/60, 103/165, 5/8, 13/20, 79/120, 2/3, 89/132, 15/22, 7/10, 17/24, 63/88, 3/4, 25/33, 47/60, 87/110, 19/24, 33/40, 37/44
  • 9828 PSL(2,27): 3/14, 20/91, 41/182, 3/13, 2/7, 53/182, 27/91, 5/14, 33/91, 3/7, 10/21, 263/546, 19/39, 23/42, 151/273, 13/21, 9/14, 59/91, 17/26, 5/7, 131/182, 31/42, 29/39, 11/14, 17/21, 19/21, 71/78, 41/42
share|improve this answer
    
@Jack Schmidt- its interesting!! Thanks! –  RDK Feb 13 '11 at 3:58

This is an attempt to explain Jack's answer geometrically, and to argue that $A_5$ is the only possible interesting (ie non-abelian) example.

We are interested in actions of finite simple groups on surfaces $\Sigma$. We are looking for simple groups $G$ that act on surfaces of different curvatures. The quotient $\Sigma$ is an orbifold of the same curvature as $\Sigma$, so we can just appeal to the classification of such orbifolds. (See, for instance, Peter Scott's article The geometries of 3-manifolds for a nice introduction to 2-dimensional orbifolds.)

There are only two infinite families of spherical orbifolds, having cyclic and dihedral fundamental groups. As these have no non-abelian simple quotients, we discard them. The remaining orbifolds correspond to the symmetries of the Platonic solids. The only interesting simple quotient we see here is $A_5$.

Famously there are seventeen Euclidean 2-orbifolds, and each has a free abelian subgroup of finite index, which we may as well quotient by. Now, I don't have a classification of these quotients to hand (the wikipedia article on Wallpaper Groups uses Conway notation, with which I'm not comfortable), but I'm willing to bet that, once again, no interesting simple quotients arise.

Finally, we are left with the hyperbolic case. It's a theorem that every finite group can be realised as the symmetries of a closed hyperbolic surface, so of course $A_5$ is no exception. Jack's calculation appears to show that $A_5$ acts on a small enough hyperbolic surface $\Sigma$ that the quotient is covered by (and hence must be, I suppose) a triangular orbifold.

share|improve this answer
    
@Henry Wilton - $A_5$ is the only non-abelian finite simple group acting on sphere, therefore we are getting a counterexample to (1), isn't it? –  RDK Feb 13 '11 at 4:02
    
Right - as Jack says, it's a counterexample to both (1) and (3). –  HJRW Feb 13 '11 at 4:45

The only groups where 1/l+1/m+1/n is greater than one are the cyclic groups, the dihedral groups, and the groups A(4), S(4), and A(5). Therefore, A(5) is the only simple groups where 1/l+1/m+1/n>1. As mentioned above, it is also a group such that 1/p+1/q+1/r<1, so it is a counterexample to (1) and (3).

All groups such that 1/l+1/m+1/n=1 are solvable, so therefore (2) is vacuously true.

Therefore your statement can be made into the stronger (and more accurate) statement: The only simple group with presentation $G=\langle x,y|x^l=y^m=(xy)^n=1,...\rangle$, when 1/l+1/m+1/n>1 is the group A(5)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.