# Given a set, we can always form a magma or how to think about (minimal) algebraic structures?

I've been told that algebra is about abstracting already existing structures and not the other way around. "What is the properties of the integers, can we generalize these properties in a more abstract sense?" We must have something to work with and not just making up axioms which nothing satisfies. According to a lecturer I had, working with abstract "objects" with no connection to existing structures is pretty much just set theory.

Anyhow.

If we flip the question; when does a set have algebraic structure? I'm aware that this the wrong way to think about it but bear with me for a second. To make this question a bit more precise I define a set to have a minimal algebraic structure if we have at least one well defined binary operator on the set. So my question becomes:

When does a set have a at least one well defined binary operator?

My naive and uninformed attempt at tackling this problem:

Given a nonempty set $S$, we can always form a minimal algebraic structure on $S$, that is $S$, can always be a magma:

Let $S$ be a nonempty set. Define the following binary operator on $S$:

$\phi: S \times S \to S$ such that $a\phi b = a, \forall a,b \in S$

Even though $\phi$ is very boring and kind of trivial, it works, right?

For example let's consider $P= \left\{p \in \mathbb{Z} \colon p \text { is prime} \right\}$ and $I = \mathbb{R} \setminus \mathbb{Q}$, two sets which at least in my mind seems do be fairly unalgebraic, but my $\phi$ is well defined on both? If this holds then $P$ and $I$ are in fact two (very trivial) magmas, so algebraic structure exists or rather can be defined onin these sets.

Can I have a set in which my $\phi$ is not well defined?

Is this nonsense? (in the sense: is this outright just wrong)

• I'm sorry if this a fuzzy question. I'm not sure I perhaps should have added the soft question and or the elementary set theory tag. – John Smith Dec 15 '13 at 13:17

Your $\phi$ defines on the set the structure of a semigroup (it is a well-known semigroup of left zeroes).

Another "quasi-minimal" example: fix an element $a$ of the set and define $xy=a$ for all $x,y$ ("quasi" since we have to fix). Then we get a semigroup with zero multiplication.

It seems that only "minimal" magma is a semigroup.

• So any set can be a semigroup? – John Smith Dec 15 '13 at 13:32
• You can put a semigroup structure on any set. The way that algebraic structures like semigroups are defined means that you can use any set you want for elements, modulo cardinality concerns. Obviously some sets make things easier to say than others. – Charlotte Aten Jun 20 '16 at 3:01

Can I have a set in which my ϕ is not well defined?

No, I think your $\phi$ allows one to put a trivial magma structure on every non empty set.

Is this nonsense?

I don't see how this could be useful, in any way.

• I'm well aware of the fact that this is not insightful or useful. I just made it up to aid my understanding of the existence of algebraic structures. I apologize though. – John Smith Dec 15 '13 at 13:34
• Why should you apologize? I just answered your question w.r.t. my knowledge. It may be that I'm wrong and this is indeed useful. Let's wait for the experts' opinions ;-) – Abramo Dec 15 '13 at 13:37