free product of groups

Given groups G iG_i, iIi\in I, their free product G 1G 2= iG iG_1 \star G_2 \star \ldots = \star_i G_i is their coproduct in Grp.

Given presentations for the G iG_i, it is straightforward to find a presentation of iG i\star_i G_i; if G i=S i|R i=F i/N iG_i = \langle S_i|R_i\rangle = F_i/N_i, where each F i=S iF_i=\langle S_i\rangle is the free group on the set S iS_i and N iF iN_i\subset F_i is the normal subgroup of F iF_i generated by the subset R iF iR_i\subset F_i, then the free product

iG i:= iS i| iR i=( iF i)/ iN i\star_i G_i := \langle \coprod_i S_i | \coprod_i R_i \rangle = (\star_i F_i)/\langle\cup_i N_i\rangle

is presented by the disjoint unions of the S iS_i and the R iR_i. As with anything satisfying a universal property, the result (up to a unique coherent isomorphism) does not depend on the presentations.

The fact that free products always exist now follows from the fact that any group has a presentation; we can always take S iS_i to be the underlying set of G iG_i and take R iR_i to be the set of all words in F iF_i that equal the identity in G iG_i. The value of the more general construction above is that one often has much smaller S iS_i and R iR_i to work with. Even if G iG_i is infinite, the S iS_i and R iR_i might be finite (in the strictest sense), making this part of finite mathematics and directly subject to the methods of combinatorial group theory.

Revised on October 8, 2010 15:17:46 by Tim Porter (