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To define a group $(G,\cdot)$ one can use the requirements:

  1. Closure
  2. Associativity
  3. A (two-sided) identity element such that $g\cdot e = e\cdot g = g$
  4. A (two-sided) inverse for each g such that $g\cdot g^{-1} = g^{-1}\cdot g = e$

We were discussing the necessity of associativity when requiring two-sided identity and inverses. I did not manage to proof associativity assuming 1, 3 and 4, but could not find a counterexample that satisfies 1, 3 and 4, while not satisfying 2. So hence the question:

Is there a non-associative multiplicative closed set, with two-sided inverses and a two-sided identity?

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Octonionic multiplication is neither commutative:

$$ e_{i}e_{j}=-e_{j}e_{i}\neq e_{j}e_{i}\ $$ if $i , j$ are distinct and non-zero,

nor associative:

$$ (e_{i}e_{j})e_{k}=-e_{i}(e_{j}e_{k})\neq e_{i}(e_{j}e_{k}) $$ if $i , j , k$ are distinct, non-zero and $e_i e_j ≠ ± e_k$ .

The existence of a norm on $O$ implies the existence of inverses for every nonzero element of $O$. The inverse of $x ≠ 0$ is given by

$$x^{-1}={\frac {x^{*}}{\|x\|^{2}}}.$$

https://en.wikipedia.org/wiki/Octonion

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  • $\begingroup$ Thanks, very interesting! $\endgroup$ – Ewoud Sep 12 '19 at 12:25

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