Isomorphisms of Graph Groups

Given a graph X, define the presentation PX to have generators the vertices of X, and a relation xy = yx for each pair x, y of adjacent vertices. Let GX be the group with presentation PX, and given a field K, let KX denote the K"-algebra with presentation PX. Given graphs X and Y and a field K, it is known that the algebras KX and KY are isomorphic if and only if the graphs X and Y are isomorphic. In this paper, we use this fact to prove that if the groups GX and GY are isomorphic, then so are the graphs X and Y. Given a graph X with vertex set V(X), we define the presentation PX to be that having as generators the elements of V(X), with a defining relation vw = wv for each pair v and w of adjacent vertices of X. PX can be regarded as a presentation of a group GX, or of a /f-algebra KX over a field K. In [1], Kim, Makar-Limanov, Neggers, and Roush proved that if the algebras KX and KY are isomorphic, then so are the graphs X and Y. (In their formulation, two generators commute provided they are not adjacent in the graph. This is sufficient for our purposes, since if two graphs are isomorphic, so are their complements.) Let K be a field. In this note we will show that if the groups GX and GY are isomorphic, then so are the algebras KX and KY, thus demonstrating: THEOREM. // the groups GX and GY are isomorphic, then so are the graphs X and Y. Let /: GX —♦ GY be an isomorphism. Denote by G2X the quotient group GX/[GX, {GX)'). Then / induces an isomorphism f2 : G2X -» G2Y. Let V{X) and V(Y) be totally ordered, and denote both orderings by <. We will not distinguish by notation between a vertex of X, the corresponding element of GX, and the image of this element in G2X. For each vertex x of X, f2{x) can be written uniquely in the form y^y^2 • • • y1¡^Cx, where Cx is an element of the commutator subgroup (G2Y)', yi < y2 < ■ ■ ■ < yn, and the integers ar are all nonzero. Define ft(x) = aiyi + a2y2 + ■ ■ ■ + anyn. We will show that the function /„ : X —> KY extends to a homomorphism /» : KX —» KY by showing that if xx' = x'x is a relation of PX, then f.{x)f.{x') = /.(a;)/*(a;') in KY. LEMMA. The commutators {[xí,Xj] | Xi < Xj and Xi and Xj are not adjacent in X} of G2X are linearly independent. PROOF. Consider the exact sequence 1 -» N -+ FX -+ GX -» 1 Received by the editors January 21, 1985 and, in revised form, May 12, 1986. 1980 Mathematics Subject Classification (1985 Revision). Primary 20F05; Secondary 20F12. ©1987 American Mathematical Society 0002-9939/87 $1.00 + $.25 per page