Proof that Fin.-Gen. Group G is Fund. Group of 2-Cell Complex

[Bacle]

Hi, Everyone:

I am looking for a proof or ref. that every finitely-generated group G is the
fundamental group of a 2-cell complex.

This is either a corollary of Reidemeister-Schreier's theorem/method for producing
a presentation of a subgroup H from the presentation of the supergroup G , or, in
some versions, it is the actual theorem.

I am not sure of how it works, but I know the following, from what I read (a paper
with missing source, unfortunately):

I know we define a right action of G on a set V, where V are the vertices --0-cells--
by adjoining vertex vi with vertex vi.g ( so that we have a regular graph whose degree
is the cardinality of G ; every vertex v has |G| outgoing vertices {v.g:g in G} , and
|G| incoming vertices ; we join vg^-1 with v through g, since vg^-1g=v=v_id); then there is a vertex between v, vg_i, vg_ig_j,...; and between vg_k and {vg_k.g_j, vg_k.g_j.g_r ,..} , etc. To each path in the graph, we associate a word: the path joining , say, vgi to vgi.gj to vgi.gj.gk is assigned the word gi.gj.gk , etc.

Then the 1-complex, i.e., the edges describe the action, and the 2-cells are
used to describe the relations of G. The relations are elements of the stabilizer of
all x, i.e., a relation is a word in G that produces a loop at each vertex v (i.e.,
for every vg_k the relation-word {g_i1.g_i2...g_ik} sends every element into a loop,
so that, for all g_o in G , we get v_go.( g_i1.g_i2...g_ik)=v_go )

We then somehow use the relations as polygons.

Thanks for any Reference.

 

[lavinia]

Hi, Everyone:

I am looking for a proof or ref. that every finitely-generated group G is the
fundamental group of a 2-cell complex.

This is either a corollary of Reidemeister-Schreier's theorem/method for producing
a presentation of a subgroup H from the presentation of the supergroup G , or, in
some versions, it is the actual theorem.

I am not sure of how it works, but I know the following, from what I read (a paper
with missing source, unfortunately):

I know we define a right action of G on a set V, where V are the vertices --0-cells--
by adjoining vertex vi with vertex vi.g ( so that we have a regular graph whose degree
is the cardinality of G ; every vertex v has |G| outgoing vertices {v.g:g in G} , and
|G| incoming vertices ; we join vg^-1 with v through g, since vg^-1g=v=v_id); then there is a vertex between v, vg_i, vg_ig_j,...; and between vg_k and {vg_k.g_j, vg_k.g_j.g_r ,..} , etc. To each path in the graph, we associate a word: the path joining , say, vgi to vgi.gj to vgi.gj.gk is assigned the word gi.gj.gk , etc.

Then the 1-complex, i.e., the edges describe the action, and the 2-cells are
used to describe the relations of G. The relations are elements of the stabilizer of
all x, i.e., a relation is a word in G that produces a loop at each vertex v (i.e.,
for every vg_k the relation-word {g_i1.g_i2...g_ik} sends every element into a loop,
so that, for all g_o in G , we get v_go.( g_i1.g_i2...g_ik)=v_go )

We then somehow use the relations as polygons.

Thanks for any Reference.

I am not sure how to do this but ... why not start with the one point join of finitely many circles, one for each generator of the group. Its fundamental group is the free group on the generators. The normal subgroup of relations in your group defines paths that must be null homotopic. Attach a 2 disc to each of a set of generators of this group and you should be done.

What needs proof it seems is:

The group of relations is finitely generated.

No extra relations are introduced.

 

[Bacle]

Thanks, Lavinia, that's a good lead.