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

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


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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.