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Let a group $G$ be defined as a $4$-tuple $G=(|G|,\mu_G,i_g,e_G)$ satisfying the following axioms:

1) $\mu_G(\mu_G(x,y),z)=\mu_G(x,\mu_G(y,z))$

2) $\mu_G(x,e_G)=x=\mu_G(e_G,x)$

3) $\mu_G(x,i_G(x))=e_G=\mu_G(i_G(x),x)$

To each group $G$, we associate a pair $G'=(|G|,\delta_G)$, with $\delta_G(x,y):=\mu_G(x,i_G(y))$. This pair characterises the group, in the sense that, if $G_1,G_2$ are two groups with $G_1'=G_2'$ then $G_1=G_2$.

Now the question: Suppose $|X|$ is any set and $\delta\colon|X|\times|X| \to |X|$ any map. I want to write down a set of axioms for the pair $X = (|X|, \delta)$, which will be necessary and sufficient for it to arise from a group $G$ in the manner described above. That is, assuming $|X|$ and $\delta$ given, find convenient necessary and sufficient conditions for there to exist a group $G$ such that $G'=(|X|,\delta)$.

My work: since $\delta$ will have to be the $\delta_G$ of some group $G$, I tried to express the group multiplication using $\delta$:

$\mu_G(x,y)=\delta(x,\delta(\delta(y,y),y))$

Then I required for this multiplication the usual axioms identifying groups, which assume the following form:

associativity:

$\delta(\delta(x,\delta(\delta(y,y),y)), \delta(\delta(z,z),z)=\delta(x,\delta(\delta(\delta(y,\delta(\delta(z,z),z)),\delta(y,\delta(\delta(z,z),z))),\delta(y,\delta(\delta(z,z),z)))$

unit:

$\delta(x,\delta(\delta(\delta(x,x),\delta(x,x)),\delta(x,x)))=x=\delta(\delta(x,x),\delta(\delta(x,x),x))$

inverse:

$\delta(y,y)=\delta(x,x)$

First question: does it make sense?

Second question: is there a better choice of axioms?

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  • $\begingroup$ The question makes good sense, and I'm pretty sure I've seen answers that involved shorter axioms than the ones in the question, but I don't remember where I saw them. (That's why this is a comment and not an answer.) $\endgroup$ Nov 1, 2017 at 16:58

1 Answer 1

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You don't need to include the $i_g$ and $e_G$ in $G$, they can be defined in terms of $\delta$ but you will need that $|G|$ is not empty. After axiom $\delta(x,x) = \delta(y,y)$, define $e_G =\delta(x,x) $ and $x^{-1}= \delta(e_G,x)$. Then the rest of the axioms become much shorter. For example the associativity axiom becomes $\delta (x,(\delta (y,z^{-1})^{-1}) =\delta (\delta(x,y^{-1}),z^{-1}). $

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