Absolute presentation of a group

In mathematics, one method of defining a group is by an absolute presentation.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:

:$G\; simeq\; langle\; S\; mid\; R\; angle.$

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 $i\; e\; 1$ for all $i\; in\; I$.

Formal Definition

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

:$langle\; S\; mid\; R,\; I\; angle.$

provided that:
# $G$ has presentation $langle\; S\; mid\; R\; angle.$
# Given any homomorphism $h:G\; ightarrow\; H$ 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 $N\; riangleleft\; G$ is a non-trivial normal subgroup of $G$ then $Icap\; N\; eq\; left\{\; 1\; ight\}\; .$

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 normal presentation is sometimes referred to as a relative presentation. The term seems rather strange as one may well ask "relative to what?" and the only justification seems to be that "relative" is habitually used as an antonym to "absolute.

Example

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

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

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

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::$langle\; a\; mid\; a^8\; =\; 1,\; a^2\; eq\; 1\; angle$

Is "not" an absolute presentation for the cyclic group of order 8 because the irrelation $a^2\; eq\; 1$ 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 $G=langle\; x\_1,x\_2\; mid\; R\; angle$ 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 $h:G\; ightarrow\; H^\{*\}$, 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.

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