> < ^ Date: Wed, 25 Jun 1997 10:59:58 +0100
< ^ From: Jim Howie <j.howie@ma.hw.ac.uk >
< ^ Subject: Re: finding relations for M23

In reply to David Joyner's message:

```Dear GAP Forum:
Following Joachim Neubueser's very interesting email reply I played
around some more with GAP. I've stumbled upon the following facts:
Let
x:=(24,21,20,13,19,5,23,3,18,16,6,9,22,14,4,2,17,8,15,12,7,11,10)
and
y:=(4,8,9,12,10)(7,6,3,5,11)(15,19,20,23,21)(18,17,14,16,22).
Then M23 is generated by x, y and
(x*y)^6=1, (x^5*y^7)^4=1.
This suggests M23 might be Free(a,b)/[(a*b)^6,(a^5*y^7)^4].
```

Could someone tell me how GAP could be used to see if this is correct?
I tried Size(...) with the maxlimit=0 option but GASMAN kicked me out
again.

```                                                                  -
David Joyner
```
```the proposed presentation Free(a,b)/[(a*b)^6,(a^5*b^7)^4] cannot
work: it presents an infinite group, by a theorem of Rick Thomas
(Cayley graphs and group presentations, Math Proc Camb Phil Soc 103
(1988), 385-387).   The conditions that ensure this are
1) that the group has a homomorphic image (in this case M23) in which
(a*b) and (a^5*b^7) really do have orders 6 and 4 respectively;
2) that 1 - # generators + sum of reciprocals of exponents on relators \le 0
(in this case 1 - 2 + 1/6 + 1/4 = -7/12).
```

In fact, since the inequality in part 2 is strict, one can go a bit
further: the kernel of the epimorphism for G = Free(a,b)/[(a*b)^6,(a^5*b^7)^4]
to M23 has a presentation with deficiency (= #generators - #relators) > 1,
so it (and hence also G) is SQ-universal, by a theorem of Baumslag
and Pride (Groups with two more generators than relators, J London Math
Soc 17 (1978) 425-426).

Jim Howie.

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