[GAP Forum] trival fp group

[GH UQ]

Dear Tim,

You asked: What I'm wondering is whether I can make GAP show me how it determined this group was trivial?

Alexander Hulpke's answer: "What GAP does is to try coset enumeration by a cyclic subgroup..." is correct. Indeed, yours is a very interesting kind of question which is addressed in some detail in:

TITLE: On proofs in finitely presented groups
AUTHORS: George Havas and Colin Ramsay
CITATION: Groups St Andrews 2005, London Mathematical Society Lecture
Note Series 340, Cambridge University Press (2007) 457-474
DOI: https://doi.org/10.1017/CBO9780511721205.010

This paper gives some specific examples and refers to:

TITLE: The Fa,b,c conjecture is true, II
AUTHORS: George Havas, Edmund F. Robertson and Dale C. Sutherland
CITATION: Journal of Algebra 300 (2006) 57-72
DOI: http://dx.doi.org/doi:10.1016/j.jalgebra.2006.01.017

which addresses some specific problems and refers to:

TITLE: Behind and beyond a theorem on groups related to trivalent graphs
AUTHORS: George Havas, Edmund F. Robertson and Dale C. Sutherland
CITATION: Journal of the Australian Mathematical Society 85 (2008) 323-332
DOI: http://dx.doi.org/doi:10.1017/S1446788708000852

which gives more information and refers to

Sutherland, D. C., ‘Computer-assisted proofs and the F a,b,c conjecture’, PhD Thesis, University of St Andrews, 2005.

In that thesis, Dale Sutherland describes a GAP package version of PEACE which she developed. However that package is no longer readily available. However PEACE is, as mentioned by Dima Pasechnik in this thread.

Colin Ramsay and I have now investigated your specific group with PEACE. The best proof that we have so far found has the following proof-word, which comes from manipulating the output from a coset enumeration over the subgroup generated by x in G = < x,y | yx^3 = x^2y; y^3x = xy^2 >. Items in square brackets, [·], are subgroup generators; items in round brackets, (·), are relators (perhaps inverted or cycled); and the remaining items are conjugating group generators.

(yyxYYXy) Yx(yyxYYXy)Xy [x][x] (XXYxxyX) xxx(XXYxxyX) xY(XXXYxxy)yX(xYYXyyy)XXX (xxxYXXy)

The way to read this is described in the first paper, extracts from which say:

"Given a successful coset enumeration of the cosets of the subgroup H in the group G, we may claim that this proves that some word ω is in H.

"A fully expanded proof-word consists of a product of subgroup generators and of conjugates of group relators (by group generators).  The subgroup generators appear as given in the presentation for H, or as the formal inverses thereof. There is no such requirement on relators for G, and they or their formal inverses may be cycled in proofs. By construction, ω and the proof-word are equivalent and, since conjugates of relators are trivial in G, the proof-word is also equivalent to a product of subgroup generators. Thus, free reduction of the proof word produces ω, while reduction after cancelling the conjugates of relators produces a product of subgroup generators."

Following this process we can see that this proof-word says that yyy = xx in G. That G is trivial follows easily.

Best wishes...  George Havas   http://staff.itee.uq.edu.au/havas