This is a thing I have been thinking on and gotten a bit frustrated so I share my thoughts here in hope for clarification.

Let $M$ be a magma, that is a set with an underlying binary operation which we denote $\cdot$. The binary operation is not necessarily associative, that is we do not necessarily have $$a\cdot (b\cdot c)=(a\cdot b)\cdot c$$ Here is the issue at hand, in group theory we can "make" a group abelian by taking the quotient with its commutator subgrouop, $G/[G,G]$, which is abelian, I am wondering if there is something similar for magmas but for associativity.

Of course having dealt with universal algebra, semirings and semigroups (which is the target here in a way) what I need to work with is a relation, more specificly a congruence relation so I figured I would start defining it as such. So I started by saying that $e \mathrel{R} f$, or $(e,f)\in R$, if there exists some $a,b,c\in M$ such that $(a\cdot b)\cdot c = e$ and $a\cdot (b\cdot c) = f$, seemed like a good place to start for me in my quest, only that I realized that there is not necessarily a unit in a magma so we do not have $x \mathrel{R} x$ which is a requirement, well let's just throw those in too then I thought. Which is the reflexive closure of $R$, that is $\text{Cl}_\text{ref}(R)$ from before.

Next I thought about transitivity which is required and I got absolutely nowhere there in my attempts primarily because I could not find anyway to proceed after setting up the equalities with elements and all. I pretty much felt it was impossible so I figured "Let's just do the congruence closure and call it a day" $\text{Cl}_\text{cng}(R)$. Which would of course be a congruence by the very definition but I feel it's a bit "cheap" so to speak. And quite frankly I am not entirely certain that it will yield satisfactory results. So the question is more "is there a way to make a magma associative that is better than this?".

  • $\begingroup$ If $S$ is a semigroup (i.e., associative magma) and $f:M \longrightarrow S$ is a magma homomorphism, then the quotient of $M$ by the kernel congruence ${\rm ker} \, f$ should be a semigroup. So ${\rm ker} \, f$ should contain a "minimal" subcongruence $\mathcal{A}$ for which $M/\mathcal{A}$ is a semigroup. I wonder if you had some examples to work through if you could tease out what the definition of $\mathcal{A}$ ought to be? $\endgroup$
    – PeterJL
    Oct 5, 2016 at 4:18
  • $\begingroup$ I had no examples to work through as I have found little on magmas and is still trying to find examples of them that fit the minimal requirements, non-associative and such. And while your statement is true it is just the universal algebra statement of congruence from my idea I feel. $\endgroup$ Oct 5, 2016 at 4:21
  • $\begingroup$ Quotienting by the congruence you defined gives a magma homomorphism $q: M \to M/\text{Cl}_{\text{cng}}(R)$ where $M/\text{Cl}_{\text{cng}}(R)$ is a semigroup ,and using PeterJL's comment we see that every such homomorphism $f:M\to S$ factors via a unique semi-group homomorphism $\bar {f}: M/\text{Cl}_{\text{cng}}(R) \to S$ such that $\bar{f} q = f$. What more would one want? $\endgroup$
    – Nex
    Oct 5, 2016 at 5:29
  • $\begingroup$ @Nex that was my thought but i feel it is "cheap" to just use the congruence closure to acquire it. $\endgroup$ Oct 5, 2016 at 5:33
  • $\begingroup$ One can probably give an explicit description of this congruence, but I personally would only be interested in finding such if I needed it to answer some question. For instance if one wanted to try and answer the question: What is a necessary and sufficient condition on $M$ such that $M/\text{Cl}_{\text{cng}}(R)$ is a one element semigroup. $\endgroup$
    – Nex
    Oct 5, 2016 at 5:47

2 Answers 2


Well, I think the way you do it is correct, but maybe you could describe it in a slightly simpler way. Just consider the semigroup presented by $\langle M \mid \{mn = m \cdot n \mid m, n \in M\}\rangle$ . In other words, take the quotient of the free semigroup on the alphabet $M$ by the relations induced by the multiplication in $M$.

EDIT. Here is an example. Consider the following magma: $$\begin{array}{c|ccc} \ast &a&b&c \\ \hline a & b&b&c \\ b & a&b&c \\ c & c&c&c \end{array}$$ Now form the semigroup with generators $\{a, b, c\}$ and relations $\{aa = b, ab=b, ac=c, ba=a, bb=b, bc=c, ca=c, cb = c, cc=c\}$

Here is the corresponding GAP session:

gap> f := FreeSemigroup("a","b","c");; 
gap> a := GeneratorsOfSemigroup( g )[ 1 ];; 
gap> b := GeneratorsOfSemigroup( g )[ 2 ];; 
gap> c := GeneratorsOfSemigroup( g )[ 3 ];; 
gap> t := f / [[a*a,b], [a*b,b], [a*c,c], [b*a,a], [b*b,b], [b*c,c], [c*a,c], [c*b,c], [c*c,c]];
gap> Size( t );
gap> rws := ReducedConfluentRewritingSystem(t);
Rewriting System for Semigroup( [ a, b, c ] ) with rules
[ [ a^2, a ], [ a*c, c ], [ c*a, c ], [ c^2, c ], [ b, a ] ]
  • $\begingroup$ Would you care to elaborate? I see you used the dot on one side but not theo ther, any reason for it? $\endgroup$ Oct 6, 2016 at 11:48
  • $\begingroup$ Yes. The dot on the right hand side is the operation of the magma $M$. On the left hand side, $mn$ is the product of the two generators $m$ and $n$. Let me know if you need a concrete example. $\endgroup$
    – J.-E. Pin
    Oct 6, 2016 at 12:55
  • $\begingroup$ Let's do an example to see if I understand it. $\endgroup$ Oct 7, 2016 at 2:24
  • $\begingroup$ What is that GAP session stuff? $\endgroup$ Oct 13, 2016 at 3:02
  • $\begingroup$ Follow the link I gave to GAP. GAP is a system for computational discrete algebra, with particular emphasis on Computational Group Theory. GAP provides a programming language, a library of thousands of functions implementing algebraic algorithms written in the GAP language as well as large data libraries of algebraic objects. Moreover, it is free and works on any computer. $\endgroup$
    – J.-E. Pin
    Oct 13, 2016 at 9:52

The congruence $R$ (which makes the magma $M/R$ associative) can be described as the transitive closure of the relation $S$ defined as follows: We have $(a,b) \in S$ if and only if there are elements $x_1,\dotsc,x_n$ in $M$ (which are not assumed to be distinct) such that both $a$ and $b$ can be written as nested products of these elements in this order. For example, the two elements $$((x_1 x_2) x_3) (x_4 (x_5 x_6))$$ $$x_1 (x_2 ((x_3 x_4) (x_5 x_6)))$$ are equivalent with respect to $S$.

The transitive closure means that $(a,b) \in R$ holds iff there are elements $y_1,\dotsc,y_n$ in $M$ such that $a=y_1$, $b=y_n$ and $(y_i,y_{i+1}) \in S$ for $1 \leq i < n$.

Since $S$ is reflexive and symmetric, $R$ is an equivalence relation. It is trivial to verify that $(a,b) \in S$ implies $(ca,cb) \in S$ and $(ac,bc) \in S$ for all $a,b,c \in M$. This property extends to $R$ (by induction on the length $n$ in the definition of $R$). Hence, $R$ is a congruence.

It is important to take the transitive closure here: It may happen that there are elements $x_1,x_2,x_3,x'_3$ in $M$ such that $(x_1 x_2) x_3 = (x_1 x_2) x'_3$ in $M$ but not $x_1 (x_2 x_3) = x_1 (x_2 x'_3)$ in $M$, and then $x_1 (x_2 x_3)$ and $x_1 (x_2 x'_3)$ will be equivalent with respect to $R$, but not necessarily with respect to $S$.

  • $\begingroup$ Why should $x_1 (x_2 x_3)$ and $x_1 (x_2 x'_3)$ be equivalent with respect to $R$? $\endgroup$
    – J.-E. Pin
    Dec 14, 2016 at 11:54
  • $\begingroup$ Certainly easier than the previous, would you care however to make a proof that it is a proper congruence relation? The closure is evidently transitive and reflexive etc, but the final one isn not evident. $\endgroup$ Dec 14, 2016 at 12:35
  • $\begingroup$ @J.-E.Pin because if they aren't it lacks transitivity. $\endgroup$ Dec 14, 2016 at 12:35
  • $\begingroup$ $x_3$ and $x'_3$ might be distinct, so I don't see your point. $\endgroup$
    – J.-E. Pin
    Dec 14, 2016 at 12:38
  • $\begingroup$ @ZelosMalum: Which step is precisely not clear to you? $\endgroup$
    – HeinrichD
    Dec 14, 2016 at 14:35

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