According to this wikipedia page, a monoid is defined as an object that contains

  1. An associative binary operation
  2. An identity element

There is no mention of the object necessarily containing a set.

The same page includes the statement "a monoid is a semigroup with an identity element". It is my understanding that a semigroup is a magma with additional constraints, is this correct?

The page also compares a monoid and a magma, stating a monoid simply has more constraints. But doesn't a monoid lack magma's constraint of having to contain a set?

So does this mean a monoid does necessitate a set? Or have I misunderstood something?

I might just be nitpicking the specific wording of a wikipedia page, but I don't want to assume it's an error before I understand it for sure.

Thanks in advance!

  • 2
    $\begingroup$ Have you looked at the definition section of the article as opposed to the definition given in the introduction paragraph? $\endgroup$ – benguin Sep 12 '16 at 17:58
  • 1
    $\begingroup$ Monoids, magmas, semigroups, groups, etc. all are supposed to be sets in the first place (or contain sets, if you mean the word "contain" in the sense of "a set is part of their specification"). $\endgroup$ – darij grinberg Sep 12 '16 at 17:58
  • $\begingroup$ How no? "Suppose that S is a set and • is some binary operation..." $\endgroup$ – Billy Rubina Sep 12 '16 at 17:59
  • $\begingroup$ It says that a monoid is an algebraic structure, which on its turn is a set with one or more finitary operations. Yes a monoid can be looked at as a magma with special properties. $\endgroup$ – drhab Sep 12 '16 at 17:59

In the magma wikipedia page, there is the following diagram:

enter image description here

So, yes. A monoid is a magma with associativity and identity which is also a semigroup with identity. Both the magma and monoid have an attached set, look both pages and see the definitions.


An algebraic structure is an object of the form $(S, f_1,\dots,f_k)$ where $f_1,\dots,f_k$ are operations on $S$, that is, they are functions $f_i\colon S^{r_i}\to S$, and $r_i$ is called the arity of the function $f_i$. A $0$-ary function is understood as a constant symbol. It is often required that $S$ be a nonempty set, but this is in many cases only a matter of convention.

So, a monoid is a structure $(S,f,id)$ where $f\colon S^2\to S$ and $id$ is $0$-ary (it is called the identity element in this context. Moreover we require that $f$ and $id$ satisfy $f(id,x)=x$ and $f(f(x,y),z)= f(x,f(y,z))$.

A magma is a structure $(S,f)$ with no additional restriction except that $f$ is $2$-ary. So even if properly speaking a monoid is not a magma, if you forget that the monoid has an identity element it becomes a magma (we call this "forgetting" operation a reduct in model theory).

Similarly, a semigroup is a structure $(S,f)$ where $f$ again needs to be associative. No identity element here, but every monoid can be understood as a semigroup by forgetting the identity element.

  • $\begingroup$ What does n-ary mean? $\endgroup$ – Samy Bencherif Sep 13 '16 at 2:45
  • $\begingroup$ A function $f\colon S^n\to S$ is said to have arity $n$, or we say that it is $n$-ary. $\endgroup$ – zarathustra Sep 13 '16 at 6:37
  • $\begingroup$ Thank you, this also clarifies the term "finitary", which means the function takes a finite number of arguments $\endgroup$ – Samy Bencherif Sep 13 '16 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.