From: neumann@vax.ox.ac.uk
To: GROUP-PUB-FORUM@maths.bath.ac.uk
Sun, 5 Mar 95 14:25:27 GMT

Dear All,

I have had a number of helpful and interesting replies to the
query I posted a couple of weeks ago about the group with
presentation $\langle a, b : ba = a^2b^2 \rangle$.  Since it
appears that not all those replies went through the FORUM, and
since it appears that some people would be interested in them,
here is a summary.  

First, bibliographic information. 
================================

Tony Crilly@Middlesex  
had drawn my attention to Cayley's paper `A problem in
partitions' {\it Messenger of Mathematics}, VII (1878), 187--188
(reprinted as paper number 722 in Cayley's Collected Papers, Vol.
11) and it was this that prompted my question.   

Dave Johnson@Nottingham 
refers to the group as the Gieseking group.  He draws attention
to Wilhelm Magnus, {\it Noneuclidean tesselations and their
groups} (1974), pp 153--156, who refers to Gieseking's `nearly
inaccessible' `privately printed Ph.D. thesis (instigated by M.
Dehn)' of 1912.  Dave also mentions his own paper on growth
series in the Murray Macbeath Conf. Proc., CUP Lecture Notes,
c.1992.  

Mike Newman@Canberra
has information about a work by Anna-Lisa Arrhenius-Wold.  A
brief paper by her is referenced by Magnus, but this is a
monograph entitled `Finite groups with two generators satisfying
the relation BA = A^2b^2', 64pp, published 1971; .  

Then some mathematical information.
==================================

Dave Johnson@Nottingham
says that `it is'nt isomorphic to the similar group with relation
a^2b^2 = ba^-1'.

Steve Linton@St Andrews
says that it is automatic `with reasonably small automata'.

Jim Howie@Edinburgh
observes that the commutator subgroup is free of rank 2, and that
the group is the split extension of the free group on $x, y$ by
the automorphism $x \mapsto y \mapsto xy$. 

Derek Holt
says that `it maps onto an extension of L(2,16) by a field
automorphism of order 2, and onto A_5 wreath C_2.'  He also
writes that its subgroups of finite index have zillions of maps
onto any small finite group you might think of, and suggests that
it may have virtually free quotients.  

Chuck Miller@Melbourne
makes the same observation as Jim Howie (above).

Walter Neumann@Melbourne
infers from the Jim Howie & Chuck Miller observation that the
group is the fundamental group of a (non-orientable) hyperbolic
3-manifold which fibres over $S^1$ with fibre the punctured
torus.  So it is not word hyperbolic.  `But it is biautomatic,
since any geometrically finite hyperbolic groups is (ECHLPT).'

Bestvina, reported by Mike Newman@Canberra
also said that one can prove the group to be automatic by
geometric arguments. 


Warm thanks to all contributors. 

All best wishes, $\Pi$eter Neumann
Queen's: 5.iii.95