I believe that magma isomorphism is defined as $\phi(x*y)=\phi(x)*'\phi(y)$. The automorphism group is the set of bijective isomorphisms from the elements of the magma to itself, under the operation of function composition. I take a magma to be just a set $X$ together with any operation from $X^2$ to $X$.

The reason I suspect that this might be the case is that:

  1. I know that groups are closely related to permutations, which is exactly what automorphisms are. If a permutation leaves something invariant, then it is not hard to generally show that its inverse does so too. The axioms of closure and inverse are met, and the requirement of identity is obviously met. Function composition is always associative, and so these functions do indeed form a group.
  2. These groups seem to potentially have the quite complex structure, and so few visible constraints that it just seems reasonable that any group could be like the automorphism group of a magma, but I don't believe that I am experienced enough in the ways of abstract algebra yet to prove this.

I'm just looking for a better way for myself to think of groups in general, that's all.


  • $\begingroup$ Regarding terminology: an isomorphism is always bijective. The word you are looking for in your first paragraph is homomorphism, or just morphism. It is then true that an automorphism is a bijective homomorphism from a magma to itself. $\endgroup$ Dec 21 '20 at 18:40
  • $\begingroup$ @CaptainLama I don't think I meant homomorphism (they might not be able to form a group under composition b/c they're not invertible, right??). I just put in that it's bijective because people here always like to know your current understanding/what you have already tried :) $\endgroup$ Dec 21 '20 at 18:55

Yes, and in fact every group is the automorphism group of some commutative monoid; see Hagen von Eitzen's answer here. His construction proceeds by showing first that every group is the automorphism group of some graph, then constructing a commutative monoid out of the graph with the same automorphism group; it's quite a nice construction.

The first sentence of the abstract of this paper claims that every group is the automorphism group of a lattice, which is more or less an idempotent commutative monoid (depending on the precise definitions).

  • $\begingroup$ Wow! Thank you so, sooo much, that's so amazing, that groups and symmetry can kind of be defined purely in terms of automorphisms and such! Also that it's even true for something as restrictive as a commutative monoid or lattice is amazing! I've always found that groups are conceptually hard because they lie at a slightly unnatural level of abstraction - $\endgroup$ Dec 21 '20 at 20:56
  • $\begingroup$ they're not as specific as a single algebraic structure, nor are they as general as a magma or something (which may be even more understandable to the general public). I don't know, but I would prefer groups be introduced more "abstractly" with automorphism than with the axioms (which could sort of be 'derived' from this, I guess, depending on what you mean exactly by symmetry and derive). :) :) :) $\endgroup$ Dec 21 '20 at 20:56
  • $\begingroup$ A while ago I wondered if every symmetry/invariant was also basically just a permutation on a set that preserved a/some functions $f(\pi(x),\pi(y))=f(x,y)$ where $\pi$ is a permutation, which I thought about after reading this: kconrad.math.uconn.edu/math216/whygroups.html and also to the link about the erlangen program that I think (?) exists. I guess that this was basically asking the same thing, where the function 'f' like in above is basically just the magma operation. $\endgroup$ Dec 21 '20 at 21:07
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    $\begingroup$ @Benjamin: as for your question about symmetries, I haven't gotten to writing this up but there's a very nice answer to a version of that question: if $G$ is a finite group acting on a finite set $X$, it is always exactly the automorphism group of a relational structure on $X$. It's a little annoying to describe what a relational structure is but concretely it's a collection of subsets of $X^n$ for various $n$, which for the action of $G$ you can take to be the $G$-orbits. Finiteness is essential here, though, unlike in the other results. $\endgroup$ Dec 21 '20 at 23:22
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    $\begingroup$ @Benjamin: you can just look at the construction in Hagen’s answer and check that it produces a commutative monoid. It’s a strictly stronger result than what you asked for. Basically the lesson is that commutative monoids are still extremely general structures. $\endgroup$ Dec 22 '20 at 18:17

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