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I am not sure the importance of A being unary here. If anyone can explain if the following is the right idea.

So let A be some unary algebra, to be specific not necessarily a mono-unary algebra. I mean that A may have more than one operation but all operations are unary. so if $\sigma$ is an automorphism on A then $\forall a \in A$ we have: $\sigma(a) = b, b \in A$ and for all $f$ such that it is an operation on $A$ $\sigma(f(a)) = f(\sigma(a))$ additionally, $\sigma$ is one-to-one and onto. then the group $\langle Aut(A), \circ, id \rangle = \{ \sigma : \sigma $ is an automorphism on $A \}$ where $\circ $ is composition and $id$ is the identity automorphism.

So if $G$ is some group isomorphic to $Aut(A)$ then we have a map $X$ where

$X : G \to Aut(A)$ such that it is homomorphic and bijective. It seems like something such as left translation (left multiplication) is the key. i.e. that the unary operations on A are analogous to left translation (or similar) in G.

oliverjones
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1 Answers1

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What you've written so far shows that you're thinking about the problem the right way, but you'll have to do more work to come up with the construction of $A$ and the proof that it works. You're absolutely right that left translation is important. See below for a full solution to the problem.

Let $G$ be a group with identity $e$. Consider the unary signature $\{f_g\mid g\in G\}$. Let $A = G$, and interpret the unary operations as right translations: for $a\in A$, $f_g(a) = ag$.

There is a map $\varphi\colon G\to \mathrm{Aut}(A)$, given by left translation: For $h\in G$, $\varphi(h) = \sigma_h$, where $\sigma_h(a) = ha$. Let's check that $\sigma_h$ is an automorphism of $A$: It is a homomorphism, since for all $g\in G$, $\sigma_h(f_g(a)) = hag = f_g(\sigma_h(a))$, and it is invertible with inverse $\sigma_{h^{-1}}$.

Now $\varphi$ is a homomorphism, since $\varphi(gh)(a) = \sigma_{gh}(a) = gha = (\sigma_g\circ \sigma_h)(a) = (\varphi(g)\circ \varphi(h))(a)$. It is injective, since if $g\neq h$, then $\varphi(g)(e) = ge = g \neq h = he = \varphi(h)(e)$. For surjectivity, let $\sigma\in \text{Aut}(A)$. I claim that $\sigma = \varphi(\sigma(e))$. Indeed, for any $a\in A$, $$\sigma(a) = \sigma(ea) = \sigma(f_a(e)) = f_a(\sigma(e)) = \sigma(e)a = \varphi(\sigma(e))(a).$$

Alex Kruckman
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  • Indeed, a permutation of $G$'s underlying set which commutes with all left multiplications by $g\in G$ is precisely a $G$-set automorphism. More generally, the group of $G$-set automorphisms of the left coset space $G/H$ is canonically isomorphic to the quotient $N_G(H)/H$. – anon Sep 19 '16 at 02:56