COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

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N. S. Mendelsohn
for example, in [2] pp. 86-95). Also by Benson and Mendelsohn [l] the
Schreier-Reidemeister generators can be expressed as words in the originally
given generators of H. We now start a second coset enumeration using the
Schreier-Reidemeister generators as the defining generators for H. This
enables us to write the originally given generators of H as words in the
Schreier-Reidemeister generators. Now, by the use of the lemma, we are
in a position to write defining relations for H in terms of its originally given
generators.
Remark. It appears that we have given an extremely roundabout procedure
for obtaining defining relations for H in terms of its given generators.
Why introduce the Schreier-Reidemeister generators at all?
The following appears to be a plausible direct procedure. Every relation
in G can be written as a product of conjugates of the given relations Ri = 1.
Hence, the group H inherits as relators the conjugates of Ri when expressed
as words in the generators of H.
It would appear that it is sufficient to take as conjugating elements one
from each coset of H. Hence H inherits the relators 0~7~ RioI where Ri
ranges over the defining relators of G, oj ranges over a set of coset representatives
and o,:~R+s~ is expressed as a word in H.
The following counter example shows that these inherited relators are
not necessarily a set of defining relators for H. The group was studied by
Baumslag and Solitar.
Let G = {A, X : Xe1A2X = A3}. Let H be the subgroup generated by X
and As. By Benson-Mendelsohn [l], H = G and in fact
Calling the right side of this equation Wit is seen that in terms of Xand As,
the group G inherits the relation XM2VX = W3. Also, since G has only
one coset and one defining relation no more than one relation can be obtained
from coset enumeration. However, G. Higman has shown that in terms
of X and As the group G requires two defining relations. Hence the extra
relation (in this case A8 W --8 = 1) cannot be deleted.
REFERENCES
1. C. T. BENSON and N. S. M ENDELSOHN: A calculus for a certain class of word problems
in groups. J. Combinatorial Theory 1(1966), 202-208.
2. W. MAGNUS, A. KARRASS and D. SOLITAR: Combinatorial Group Theory (Interscience
Publishers, New York, 1966).
3. N. S. MENDELSOHN: An algorithmic solution for a word problem in group theory.
Canad. J. Math. 16 (1964), 509-516; correction 17 (1965), 505.
Nielsen transformations
M. J. DUNWOODY
LET G be a group with n generators. Let Z be the set of ordered sets of n
generators of G.
If x is a permutation of the set { 1, 2, . . . , n} then a, will denote the permutation
of Z such that
kl, . . . , gJan = (gl,, g2=, . . . , h-J.
If i&(1,2, . . .) n}, i 9 j, then a-i, ai:. will denote the permutations of ,Z
such that
kl, - - .Y g&-i = (gl, g2, . * ~3 gi-19 gi19 gi+l, . . -9 gJ,
(g1, . . . , &)ai:j = (gi, * * *v gj-13 gigj, gj+l, . . *, cl).
Let A be the group of permutations of Z generated by all the above.
It is sometimes useful in group theory to know the transitivity classes of
C under A. Let F be the free group on generators xl, x2, . . ., x,,. If
kl, 82, . . . , g,), (hl, h2, . . ., h,) belong to Z and R, S are the kernels of the
respective homomorphisms 8, 4 of ,F onto G such that Xi0 = gi, Xi4 = hi,
i= 1, . . . . n, then there is an automorphism y of F such that Ry = S if
(a, g2, . . ., gn>, @l, ha, . . . , h,) belong to the same transitivity class of Z
under A. When such an automorphism of F exists there is for instance an
isomorphism induced between F/[R, R] and F/[S, S]; these groups need not
be isomorphic if (gi, g2, . . . , g,), (hl, h2, . . ., h,) belong to different transitivity
classes under A.
The problem I consider is the following:
If G has n- 1 generators, then in every transitivity class of Z under A is
there a set of generators one of which is the unit element?
The answer to this question is yes if G is finite and soluble; in fact one has
the following :
THEOREM. If G is a finite soluble group with n- 1 generators, then A is
transitive on Z.
Proof. The proof is by induction on the length c of a chief series of G. If
c = 0, the theorem is trivial. Assume then that cs-0, and that the result is
true for c- 1.
Suppose now that
E=M,,-=Ml-=Mz-= . ..-=M.=G
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