# Is there a term for a magma with identity (only)?

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?

• 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 8 '17 at 21:31
• Probably called 'do we really care about this?'. – Pedro Tamaroff Jan 8 '17 at 21:42
• 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 8 '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 8 '17 at 22:21
• I'd call it Magmortar. – Billy Rubina Jan 8 '17 at 23:06

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.

• Got a good source of the definition of divisibility? – Zelos Malum Jan 9 '17 at 15:25
• 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 9 '17 at 17:10
• Danke${}{}{}{}$ – Zelos Malum Jan 10 '17 at 4:01

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.