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Let $f$ be a function $f:X^2\to Y$. Let $G_f$ be a subset of the symmetric group on $X$, such that $G$ consists of all of the permutations $\sigma$ satisfying $f(\sigma(a),\sigma(b))=f(a,b)$ for the chosen $f$. $X$ need not be finite, and also $Y$ need not coincide with $X$, or even have equal cardinality to it; it is just some arbitrary co-domain.

Question:

Is every group just isomorphic to some $G_f$, if we make the correct choice of $f$? Alternatively, for any group does there exist some function $f$ such that $G_f$ if isomorphic to the group in question?

Purpose:

I'm looking for ways to motivate the group axioms by showing that they describe something where the level of abstraction is no more and no less than is required to describe this particular object.

What I already know/have figured out/have tried:

  • Any $G_f$ satisfies the group axioms. Closure) if $f(\sigma_1(a),\sigma_1(b))=f(a,b)$ and $f(\sigma_2(a),\sigma_2(b))=f(a,b)$, then the composition $\sigma_3=\sigma_1\circ\sigma_2$ or $\sigma_2\circ\sigma_1$ will also satisfy $f(\sigma_3(a),\sigma_3(b))=f(a,b)$ once decomposed, implying closure. Identity) the identity permutation $e$ trivially satisfies $f(e(a),e(b))=f(a,b)$. Inverses) if $f(\sigma(a),\sigma(b))=f(a,b)$, then $\sigma^{-1}$ will also satisfy $f(\sigma^{-1}(a),\sigma^{-1}(b))=f(a,b)$, because $f(a,b)=f(e(a),e(b))=f(\sigma(\sigma^{-1}(a)),\sigma(\sigma^{-1}(b)))=f(\sigma^{-1}(a),\sigma^{-1}(b))$. Associativity) permutations are functions, and function composition is always associative.
  • As per this question I've asked, I know that every group is the automorphism group of some magma. In the language of my question here, I think that's basically saying that if $f:X^2\to X$ then any group is isomorphic to the set of all $\sigma$ under composition such that $\sigma(f(x,y))=f(\sigma(x),\sigma(y))$, but this doesn't seem exactly the same as what I have above. I think that because my condition here is even less restrictive the automorphism group of a magma, I strongly suspect that the answer to my question is yes, but I absolutely lack the algebraic prowess to justify this hunch beyond intuition. I don't know enough about lattices to follow what's going on in the answer to this question.

Note:

If the answer to this question is affirmative, then I am planning to also - at some point in the future - use this to ask a question about how this approach to compares to the magma-automorphism approach and binary-relations approach of defining the group axioms, and which way and how these concepts actually relate to the applications of group theory, such as symmetry in physics and geometry, the abstract concept of in-variance in math, how to interpret this in the case of, say, a rubrics cube, conjugation and cosets, the isomorphism theorems, & more.

Shaun
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Pineapple Fish
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    This is basically equivalent to asking whether every group is the automorphism group of a graph (where the edges can have "colors" corresponding to the elements of $Y$), which is already established by Hagen's answer from the previous thread. – Qiaochu Yuan Dec 24 '20 at 17:23
  • @QiaochuYuan I'll revist that question with that in mind – Pineapple Fish Dec 24 '20 at 17:24
  • @QiaochuYuan ok I believe you, but I still don't really see how exactly – Pineapple Fish Dec 24 '20 at 17:28
  • @QiaochuYuan "if is a finite group acting on a finite set , it is always exactly the automorphism group of a relational structure on ... Finiteness is essential here, though, unlike in the other results." According to Hagen Von Eitzen's answers, the above constructions should ignore that finiteness condition. Do you think that viewing groups as magma-automorphisms or the $G_f$ like in the above question would be better for explaining the essence of group theoretical concepts to a beginner (assuming an understanding of isomorphism)? – Pineapple Fish Dec 24 '20 at 17:36
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    I think this is not a good way of thinking about groups; it seems to me to be including a substantial additional complication into the definition that is unnecessary. We took the abstract perspective for a reason; among other things, there are many ways to define groups that don't automatically equip them with faithful actions on anything, most notably taking various kinds of quotients, and the freedom to do that is extremely valuable. – Qiaochu Yuan Dec 24 '20 at 17:37
  • @QiaochuYuan Thanks, I'll need some time to think about this point. – Pineapple Fish Dec 24 '20 at 17:42
  • @QiaochuYuan well I guess, like which is more fruitful for spotting examples of groups in the real world and inventing your own group theoretic concepts, or doing research? I am not an empiricist, but I am also not a [strict] formalist. I believe that one can have an abstract conception of something whilst still having something that works to fall back on when you have to prove something. Surely there is a reason that we choose to study some problems related to groups rather than others, which is presumably motivated by the objects that we are studying, right? – Pineapple Fish Dec 24 '20 at 18:03
  • @QiaochuYuan Never mind, sorry, so what you said about relational structures: does this work in the infinite case: if $\sigma(x)~\sigma(y)\implies x~y$ means that $\sigma\in G$ (relative to a given arbitrary binary relation $~$), then is any group isomorphic to G under function composition if we choose the correct binary relation? On the one hand it seems to follow from the last two questions I've posted, on the other if this is the same type of thing meant by relational structure then it would only work in the finite case? Sorry.. – Pineapple Fish Dec 24 '20 at 18:59
  • This is equivalent to asking whether every group is the automorphism group of a (simple, undirected) graph, which follows from the second part of Hagen's answer. This is a different statement from the one I made about relational structures, which is not about recovering all groups but about recovering all finite groups together with all of their faithful actions on finite sets. These automorphism-of-graph etc. constructions only show that there exists some faithful action with a certain property. – Qiaochu Yuan Dec 24 '20 at 19:58
  • Let us continue this discussion in chat. – Pineapple Fish Dec 24 '20 at 22:11

1 Answers1

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This follows from the beginning of Hagen von Eitzen's answer that I linked to previously. Every group occurs in this way, and we can take $X = Y = G$ and $f : G \times G \to G$ to be the function

$$f(x_1, x_2) = x_1^{-1} x_2.$$

Then we can check that if $g \in G$ and $L_g : X \to X$ denotes the left-multiplication-by-$g$ map $x \mapsto gx$ that we have

$$f(L_g x_1, L_g x_2) = (gx_1)^{-1} gx_2 = x_1^{-1} g^{-1} g x_2 = x_1^{-1} x_2$$

so $L_g$ is an automorphism of $f$ (which defines a kind of graph structure on $G$ where the edges are directed and colored by the elements of $G$ itself; $f$ above is describing the "difference" between two elements of $G$).

Conversely, suppose $\varphi : X \to X$ is an automorphism of $f$ and set $g = \varphi(e)$. Then

$$f(\varphi(e), \varphi(x)) = f(g, \varphi(x)) = g^{-1} \varphi(x) = f(e, x) = x$$

gives that $\varphi(x) = gx$ for all $x \in X$. So $\varphi = L_g$. Hence the automorphisms of $f$ are exactly the left multiplications $L_g$.

The function $f$ above can be thought of as describing the structure of a right $G$-torsor on $G$, and it's classical that the automorphism group of a $G$-torsor is $G$ again; this is a more precise version of Cayley's theorem.

Qiaochu Yuan
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