# Monoid of specifications for a group

1. Oct 9, 2012

### Stephen Tashi

Monoid of "specifications" for a group

The question of whether there is any standard math associated with specifications of ordered pairs on a group went nowhere (https://www.physicsforums.com/showthread.php?t=640395), so I will spell out what I have in mind.

It appears possible to define a monoid of "specifications" for a group G, as sketched below. Is there a technical name for this monoid? Is it a special case of some standard structure in group theory?

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Let G be a group. Each element $g \in$ G can be identified with the function that permutes th-e elements of $G$ by left multiplication. In what follows we will consider $g$ to be the set of ordered pairs $\{(x,gx): x \in G \}$.

Define a specification $s$ on $G$ to be a set of ordered pairs of elements of $G$ such that there exists an element $g \in G$ such that $s \subset g$.

For example, Let $G$ be the group of permutations on the set of 4 letters $\{a,b,c,d\}$ then the set $s = \{(a,b),(c,a)\}$ is a specification since the group element $g_1 = \{(a,b),(b, c),(c,a),(d,d) \}$ contains $s$ as a subset. (There is also another group element $g_2 = \{(a,b),(b,d),(c,a),(d,c)\}$ that contains $s$.) In general, a specification need not define a unique group element.)

For a specification $s$ on a group $G$ , denote by $G(s)$ the set of elements in $G$ that contain $s$.

Examples:

In the previous example $G(s) = \{g_1,g_2\}$.

If $g \in G$ then $G(g) = g$ since the ordered pairs of $g$ define it uniquely.

$G\{\emptyset \}$ is the entire set of elements of $G$.

Define a multiplication operation on two specifications as follows:

Let $s,t$ be specifications on the group $G$. Define the product $s t$ to be the specification consisting the all ordered pairs ${x,y}$ such there is some ordered pair $(a,y) \in s$ and some ordered pair $(x,a) \in t$

A specification defines a 1-1 function from a subset of $G$ onto another subset of $G$. The product of two specifications amounts taking the composition of two such functions on the intersection of their domains.

The identity element $I$ of the group $G$ defines a specification that is a multiplicative identity for the above product operation. The set of all possible specifications for a group $G$ forms a monoid under the product operation.

The monoid of specifications is not the same as a monoid formed by subsets of the group, i.e., in general, $G(s t)$ need not equal $G(s) G(t)$.

2. Oct 9, 2012

### Norwegian

Re: Monoid of "specifications" for a group

Your first example does not correspond well to your definition of specification. There seem to be a mix-up between the 4 element set and its group of permutations.

Using the definition, G(s) is only one element (for nontrival s), but your example allow for some more structure.

So I suggest clearing up that confusion, and also give some motivation for why your monoid is interesting, as monoids can be made out of almost anything.

3. Oct 9, 2012

### Stephen Tashi

Re: Monoid of "specifications" for a group

Yes, it's a faulty example and faulty definition! For a specification to be non-trivial, I need to define it as a set of ordered pairs of elements of a set that is acted upon by a group rather than necessarily being a set of ordered pairs of elements of the group.

I'll try this:

Define:

A specification of a group $G$ that acts on a set $S$ is a set of ordered pairs of elements of $S$ such that no two distinct ordered pairs have the same first element.

Let $G$ be the group of permuations acting on the set of 3 integers {1,2,3}.
The 6 elements of the group $G$ are the functions
$a = \{(1,1),(2,2),(3,3)\}$
$b = \{(1,1),(2,3),(3,2)\}$
$c = \{(1,2),(2,1),(3,3)\}$
$d = \{(1,2),(2,3),(3,1)\}$
$e = \{(1,3),(2,1),(3,2)\}$
$f = \{(1,3),(2,2),(3,1)\}$

An example of a specification of the above group on the set {1,2,3} would be $s = \{(1,3)\}$

(In my small example, if I put two ordered pairs in the specification, this is enough to determine a unique group element, but with a larger set of integers and a larger permutation group, I could put two ordered pairs in a specification without determining a unique group element.)

$G(s) = \{e,f\}$ since both $e$ and $f$ contain the ordered pair $(1,3)$.

$G(b) =G(\{ (1,1),(2,3),(3,2) \}) = {b}$

I don't have any strong motivation for this. I't simply idle curiosity and the fact that I happen to glance at the book "introduction To Semigroups" by Mario Petrich because I was moving books around while rearranging my bedroom.

The standard game in finite groups is "I'll give you the order of the group and you tell me about possibilities for its subgroups". ( I don't understand why that game seems to be the only game in town- except that in practical applications, you might be able to count the symmetries of an object and want to know about the group of symmetries.) It seems to me that there might be other games involving "I'll tell you a little about the action of a group on this set and you tell me about the possibilities for the group".