3

The title of the question motivates the specific mathematical question given in the next section.


Let $A$ be finite set.

For an integer integer $n \ge 0$, a function $s: \{k \mid k \le n \land k \gt 0\} \to A$ is said to be a word (or string) in the alphabet $A$ of length $n$. The collection of all words in $A$ is denoted by $ \mathcal W$. Note that $\emptyset \in \mathcal W$; it is called the null string and is denoted by $[\,]$.

In a natural fashion, any two strings $\mathcal W$ can be concatenated,

$\quad (s,t) \mapsto s \mid t$

so that $(\mathcal W, \; \mid \;)$ is a free monoid.

The elements in $A$ can be regarded as words of length $1$ in $\mathcal W$ and by abuse of notation we write $A \subset \mathcal W$.

Let $\mathcal R \subsetneq \mathcal W$ be a finite set with $A \subset \mathcal R$ and $[\,] \in \mathcal R$.

Let $\Gamma: \mathcal W \to \mathcal R$ be a surjective mapping satisfying

$\tag 1 \Gamma ([\,]) = [\,]$

$\tag 2 \forall a \in A, \quad\Gamma (a) = a$

$\tag 3 \forall s,t \in \mathcal W, \quad \Gamma(s \mid t) = \Gamma\bigr(\Gamma (s) \mid \Gamma (t)\bigr)$

Are there any mathematical specifications (examples) where $\text{(1)-(3)}$ holds?

My Work

I am close to working out the details to representing the symmetric groups with such a framework. I could not find any references, but if it has already been done then that would be an example; any links/comments would be appreciated.

I have a working python program using this representation theory that allowed me to supply this answer to a motivational question.

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    This seems related to (strong) confluence with (an incomplete list of) ingredients $s | t \rightarrow \Gamma(s)|t$ and $s | t \rightarrow s | \Gamma(t)$. Your $\Gamma$ also seems to be strongly normalizing. In the context of both, Newman's lemma may apply. – Eric Towers Sep 27 '20 at 16:11
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    What you describe sounds like a confluent rewriting sytem. For the symmetric groupcs, the Coxeter presentations provide such a system. – ahulpke Sep 27 '20 at 16:51
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    More generally (as a follow up to @ahulpke), Coxeter groups satisfy this. – Tobias Kildetoft Sep 27 '20 at 17:19
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    As a minor quibble, I think you are looking at a free monoid (semigroup with identity) rather than a free semigroup. – Derek Holt Sep 28 '20 at 13:42
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    But the answer to your question is that all finite groups (indeed all finite monoids) have this property. A typical standard form is the lexicographically least among the shortest strings that represent that group element, but there are many others. For example, for finite solvable groups it is customary to use a standard from based on a composition series. – Derek Holt Sep 28 '20 at 14:24
  • @DerekHolt Can the algorithm be specified when there can be several 'least string' representations (multiplicity of choice). – CopyPasteIt Sep 28 '20 at 16:04
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    The principal method here is called the Knuth-Bendix completion algorithm. It can be applied to many algebraic structures (for example, it computes Gröbner bases when applied to polynomial rings), so you need to look for applications to groups and monoids. Typically, you specify an ordering on strings (with certain required properties), and then it aims to reduce strings to the least string under this ordering that represents the same group element. – Derek Holt Sep 28 '20 at 17:59

2 Answers2

1

Here we explain the underpinnings of our main example, the symmetric group $S_n$ for $n \ge 2$.

The set $A$ of generators are the transpositions.

The standard/reduced form $\mathcal R$ is exhibited by this form

$\quad \prod\limits_{k=1}^{n}\, \big(k \; \omega(k)\big) = \big(1 \; \omega(1)\big) \circ \big(2 \; \omega(2)\big) \circ \dots \circ \big(n \; \omega(n)\big)$

where for all $k$, $\;\omega(k) \ge k$ (for more details see this).

At this point we have $\mathcal R$ and $S_n$ in bijective correspondence. The key to specifying $\Gamma$ is the following observation:

Let $s$ and $t$ be two transpositions; the following cases are mutually exclusive,

Case 1: $st = ts = [\,]$.
Case 2: $st = ts \ne [\,]$.
Case 3: $st \ne ts \text{ and } \exists \text{ transpositions } s',t' \in A \text{ such that }$
$\quad \quad \quad st = t's \land st = ts'$

Symbolic processing is used to derive both $s'$ and $t'$.

I implemented $\Gamma$ with the Python programming language but it is not efficient (or pretty). The algorithm employs a type of bubble sort (multiple times) along with the symbolic processing.


Example

Here we crank out the answer after taking two 'random' (but long) $\mathcal R$ words from $S_4$; the elements are

$\quad [(12)\,(23) \,(34)] \text{ and } [(14)\,(24)\,(34)]$

Anytime we have an $\mathcal R$ word of length $3$ it will have a $(34)$ on the far right. So, we begin

$\quad [(12)\,(23) \,(34)] \text{ || } [(14)\,(24)\,(34)] = (12)\,(23) \, (13) \, (34) \,(24) \,(34) =$ $\quad \quad \quad \quad \quad \quad \quad (12)\,(23) \, (13) \, (23) = [\;]$

Well that was amusing! I simply scanned my answer here looking for the first two words of length $3$ and came up with a 'cancel out story'. I encourage the interested reader to try it out on other examples.

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1

After working on this answer concerning the group generated by $(13)$ and $(1234)$ in $S_4$, I was motivated to construct a non-commutative group of order $10$.

Here are the defining relations allowing us to construct $\Gamma$,

$\tag 1 TT = [\,]$ $\tag 2 CCCCC = [\,]$ $\tag 3 CT = TCCCC$

and here is the group table:

enter image description here

Some Comments on the Technique

It wasn't necessary to supply any logic/argument here to allow us to assert that the table satisfies the group axioms. Rather, we used a Python program to check that the operation is associative. Also, the null string is the identity and, by inspecting the table, we see that every element has an inverse.

Consider

$\quad CT = TC^k$

Setting $k = 0$ would result in $S_2$, so remove it from consideration.

Since we're not interested in the construction of an abelian group, we reject the $k = 1$ setting.

For $k \in \{2,3\}$ the constructed binary operation isn't associative if we insist there are $10$ distinct words as shown in the group table. Said another way, the relation would 'present' a smaller (collapsed) group same as when $k = 0$, but we don't really care what that group is.

So, the only way to 'seal the deal' with this idea is with $k = 4$, i.e. with $\text{(3)}$.

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