Absolute Presentation Of A Group

In mathematics, an absolute presentation is one method of defining a group.B. Neumann, ''The isomorphism problem for algebraically closed groups,'' in: Word Problems, Decision Problems, and the Burnside Problem in Group Theory, Amsterdam-London (1973), pp. 553–562.

Recall that to define a group $G$ by means of a presentation, one specifies a set $S$ of generators so that every element of the group can be written as a product of some of these generators, and a set $R$ of relations among those generators. In symbols:

:$G \simeq \langle S \mid R \rangle.$

Informally $G$ is the group generated by the set $S$ such that $r = 1$ for all $r \in R$. But here there is a tacit assumption that $G$ is the "freest" such group as clearly the relations are satisfied in any homomorphic image of $G$. One way of being able to eliminate this tacit assumption is by specifying that certain words in $S$ should not be equal to $1.$ That is we specify a set $I$, called the set of irrelations, such that $i \ne 1$ for all $i \in I.$

Formal definition

To define an absolute presentation of a group $G$ one specifies a set $S$ of generators and sets $R$ and $I$ of relations and irrelations among those
generators. We then say $G$ has absolute presentation

:$\langle S \mid R, I\rangle.$

provided that:
# $G$ has presentation $\langle S \mid R\rangle.$
# Given any homomorphism $h:G\rightarrow H$ such that the irrelations $I$ are satisfied in $h(G)$, $G$ is isomorphic to $h(G)$.

A more algebraic, but equivalent, way of stating condition 2 is:

:2a. If $N\triangleleft G$ is a non-trivial normal subgroup of $G$ then $I\cap N\neq \left\ .$

Remark: The concept of an absolute presentation has been fruitful in fields such as algebraically closed groups and the Grigorchuk topology.
In the literature, in a context where absolute presentations are being discussed, a presentation (in the usual sense of the word) is sometimes referred to as a relative presentation, which is an instance of a retronym.

Example

The cyclic group of order 8 has the presentation
:$\langle a \mid a^8 = 1\rangle.$

But, up to isomorphism there are three more groups that "satisfy" the relation $a^8 = 1,$ namely:
:$\langle a \mid a^4 = 1\rangle$
:$\langle a \mid a^2 = 1\rangle$ and
:$\langle a \mid a = 1\rangle.$

However none of these satisfy the irrelation $a^4 \neq 1$. So an absolute presentation for the cyclic group of order 8 is:
:$\langle a \mid a^8 = 1, a^4 \neq 1\rangle.$

It is part of the definition of an absolute presentation that the irrelations are not satisfied in any proper homomorphic image of the group. Therefore:
:$\langle a \mid a^8 = 1, a^2 \neq 1\rangle$

Is ''not'' an absolute presentation for the cyclic group of order 8 because the irrelation $a^2 \neq 1$ is satisfied in the cyclic group of order 4.

Background

The notion of an absolute presentation arises from Bernhard Neumann's study of the isomorphism problem for algebraically closed groups.

A common strategy for considering whether two groups $G$ and $H$ are isomorphic is to consider whether a presentation for one might be transformed into a presentation for the other. However algebraically closed groups are neither finitely generated nor recursively presented and so it is impossible to compare their presentations. Neumann considered the following alternative strategy:

Suppose we know that a group $G$ with finite presentation $G=\langle x_1,x_2 \mid R \rangle$ can be embedded in the algebraically closed group $G^$ then given another algebraically closed group $H^$, we can ask "Can $G$ be embedded in $H^$?"

It soon becomes apparent that a presentation for a group does not contain enough information to make this decision for while there may be a homomorphism $h:G\rightarrow H^$, this homomorphism need not be an embedding. What is needed is a specification for $G^{*}$ that "forces" any homomorphism preserving that specification to be an embedding. An absolute presentation does precisely this.

References

Combinatorial group theory