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

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?

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

• Thanks, very interesting! – Ewoud Sep 12 '19 at 12:25