1Cayley graphs and the word metric

IV Topics in Geometric Group Theory

1.3 Finitely-presented groups

Let’s try to consider groups that are slightly more complex than free groups. If a

group Γ is generated by

S

, then we have a surjective homomorphism

F

(

S

)

→

Γ.

Let K = ker η. Then the first isomorphism theorem tells us

Γ

∼

=

F (S)

K

.

Since we understand

F

(

S

) quite explicitly, it would be nice if we have a solid

gasp on

K

as well. If

R

normally generates

K

, so that

K

=

hhRii

, then we say

that hS | Ri is a presentation for Γ. We will often write that

Γ

∼

=

hS | Ri.

Example.

Z

2

= ha, b | aba

−1

b

−1

i

Definition

(Finitely-presented group)

.

A finitely-presentable group is a group

Γ such that there are finite S and R such that Γ

∼

=

hS | Ri.

A finitely-presented group is a group Γ equipped

S

and

R

such that Γ

∼

=

hS |

Ri.

Presentations give us ways to geometrically understand a group. Given a

presentation

P

=

hS | Ri

, we can construct space

X

P

such that

π

1

X

P

∼

=

hS | Ri

To do so, we first construct a rose with

|S|

many petals, each labeled by an

element of

S

. For each

r ∈ R

, glue a disk onto the rose along the path specified

by r. The Seifert–van Kampen theorem then tells us π

1

X

P

∼

=

Γ.

Example.

We take the presentation

Z

2

∼

=

ha, b | aba

−1

b

−1

i

. If we think hard

enough (or a bit), we see this construction gives the torus.

Conversely, if we are given a connected cell complex

X

, we can obtain a

presentation of the fundamental group. This is easy if the cell complex has a

single 0-cell. If it has multiple 0-cells, then we choose a maximal tree in

X

(1)

,

and the edges

S

=

{a

i

}

not in the maximal tree define a generating set for

π

1

X

(1)

. The attaching maps of the 2-cells in

X

define elements

R

=

{r

j

}

of

π

1

X

(1)

, and these data define a presentation P

X

= hS | Ri for π

1

X.

This is not canonical, since we have to pick a maximal tree, but let’s not

worry too much about that. The point of maximal trees is to get around the

problem that we might have more than one vertex.

Exercise. If X has one vertex, then Cay

S

π

1

X =

˜

X

(1)

.