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If we start from magmas and consider: associativity, identity, invertibility (divisibility). We will theoretically get $2^3=8$ structures by regarding whether such structure possess these properties. As is shown in the picture: https://en.wikipedia.org/wiki/Magma_(algebra)#/media/File:Magma_to_group2.svg

There miss two structures - associative, divisible magma and magma with identity. Well, we know associative quasigroup is a group. So, there is only one left. Is there a term for it?

And it is necessary to differ between invertibility and divisibility?

Upc
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    If you have no identity, then it doesn't make sense to talk about invertibility, but divisibility (left and right) still make sense. – tomasz Jan 08 '17 at 21:31
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    Probably called 'do we really care about this?'. – Pedro Jan 08 '17 at 21:42
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    In answer to your initial question, I guess you might find the terms "left unital magma", "right unital magma" and "two-sided unital magma", according to the type of identity element. A loop by definition requires the identity element to be two-sided. As Pedro says, you are probably asking about something not worth naming. – qman Jan 08 '17 at 21:45
  • @qman but these derived names with adjectives are worth existing, and exist because the nomenclature is flexible enough to produce (composite) name for this specific object. – YCor Jan 08 '17 at 22:21
  • I'd call it Magmortar. – Red Banana Jan 08 '17 at 23:06

3 Answers3

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The term "unital magma" does get used, including by Bourbaki (Algebra I, p. 12):

A magma with an identity element is called a unital magma. Here, an identity element is defined to be two-sided.

Divisibility does not imply invertibility, so they are distinct. One can have divisibility without an identity element; one needs both for invertibility. Note: a left inverse and a right inverse might not be the same, but invertibility does not demand this.

qman
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  • Got a good source of the definition of divisibility? – Zelos Malum Jan 09 '17 at 15:25
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    Bourbaki defines left and right cancellable elements (p.14) and left and right invertible elements in terms of an identity element (p. 15). But that is not what you're asking. In this context, divisiblility is as in a quasigroup means, for every pair $a$ and $b$ in the set, both existence and uniqueness of $x$ and $y$ that solve $ax=b$ and $ya=b$. Not a great word. And not the most authoritative reference, but it gives several references. It is clear that quasigroup need not have an identity element, and hence need not support invertibility. – qman Jan 09 '17 at 17:10
  • Danke${}{}{}{}$ – Zelos Malum Jan 10 '17 at 04:01
  • Can be exists a partial magma with identity element but that is not a magma? Only requiment for magma is a Set that is closed under an operation but I need to do get this: preserve identity element but without the need for the Set to be closed by definition – Jack Rock Mar 06 '21 at 18:59
  • I don't have a reference, but in the way I've seen it used, divisibility is the "existence" and cancellation the "uniqueness", so one could say perhaps that quasigroups are those magmata which are both cancellable and divisible...? – May Emerson Aug 14 '25 at 23:04
  • @MayEmerson - In the sense of "$a$ divides $b$" or "$b$ is divisible by $a$", what you say has some sense. However, that use applies to elements, and the question might be whether it would it be taken to extend to structures. – qman Aug 15 '25 at 11:48
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Here's an example of such a structure.

$$\begin{array}{c|ccc} \cdot & e & a & b \\ \hline e & e & a & b\\ a & a & b & a\\ b & b & b & a\\ \end{array}$$

We see that $e$ is an identity element. However, $(S, \cdot)$ is not associative, since $(a\cdot b)\cdot b = a\cdot b = a$, while $a\cdot (b\cdot b) = a \cdot a = b$.

Also, $(S, \cdot)$ does not have inverses. There is no element $x\in S$ such that $a\cdot x = e$.

Also, though you didn't mention commutativity, $(S, \cdot)$ is not commutative: $a\cdot b = b$, while $b\cdot a = a$. We actually could make it commutative, though:

$$\begin{array}{c|ccc} \star & e & a & b \\ \hline e & e & a & b\\ a & a & b & b\\ b & b & b & a\\ \end{array}$$

$(S, \star)$ still has an identity, still lacks inverses, and is now commutative. And $(S, \star)$ is still non-associative, since $(a\star b)\star b = b\star b = a$, while $a \star (b\star b) = a \star a = b$.

I guess one way to make a magma with identity is to start with whatever magma you want, and then add another element, $e$, and extend your binary operation to make $e$ act as an identity. So I think this might explain why the some people suggested in the comments that this type of structure might not be very interesting, and why it doesn't have its own name.

mathmandan
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Yes, but it's very obscure. At least two papers call such an object a quasimonoid.

I suspect there are a handful of other papers in existence using it, but by no means is this a standard term, and probably no one would know what you're talking about if you just said it with no explanation.

Oh, and this is essentially unrelated, but as for the objection that quasimonoids have far too little structure to deserve their own term, well, there exist far worse examples. A "U-semigroup" is a semigroup with an involution, in the most general sense possible - a function such that f(f(x)) = x. Every semigroup can be made into a U-semigroup, and a great very many don't even have a unique choice! Or much worse: a "zroupoid" is a magma with a distinguished element. Yes, really.

May Emerson
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  • The trick to making $U$-semigroups interesting is to further constrain the involution. – J.G. Jul 30 '25 at 07:57
  • @J.G. Yes, of course, and this goes for any of these other structures as well - they seem pointlessly general, but the reason they are given names is as a starting point to define less general and more interesting structures.

    (There is something to be said though about deriving a rich structure the "opposite" way - if you take some property X, even a very trivial property such as existence of an identity, you can imply quite a lot of structure by defining a very slight generalization of X, and then asking what models satisfy gen(X) but not X. I find this very interesting.)

     – May Emerson Aug 09 '25 at 23:45