Say we have an algebra $(A, +, \cdot)$, where $(A, +)$ is an Abelian Group. All we know about $\cdot$ is that it is both left and right distributive over addition. So, $\forall a,b,c \in A, a \cdot (b+c) = (a \cdot b) + (a \cdot c)$ and $(b+c) \cdot a = (b \cdot a) + (c \cdot a)$.

We can't assume $ \cdot $ is associative, commutative, or anything else besides distributive.

Do we know whether multiplication is power associative or not? That is, for all $a$, powers of $a$ are associative (e.g $a \cdot (a \cdot a) = (a \cdot a) \cdot a$).

If so, what would the proof look like? If not, is there a counterexample?

I attempted this myself, but I couldn't find any hints of a proof, so I tried to produce a counterexample, similarly with no luck.

  • $\begingroup$ @AniruddhAgarwal Induction on the size of the set? If so, I did try that, but I'll try again. If not, induction on what? $\endgroup$
    – RothX
    May 1 '19 at 16:15
  • $\begingroup$ he may have meant induction on the power of $a$ with $a^2$ being trivially true, but has deleted his comment since... don't think it is easy to prove that way $\endgroup$
    – gt6989b
    May 1 '19 at 16:16
  • $\begingroup$ Would you be able to argue something from definition of multiplication via addition, like $$a \cdot (a \cdot a) = a \cdot (a + a + \ldots + a) = a\cdot a + \ldots + a\cdot a = (a \cdot a) \cdot a?$$ $\endgroup$
    – gt6989b
    May 1 '19 at 16:18
  • 2
    $\begingroup$ Suggest you fix "ring distributive" to "right distributive" since as you spell it out that's what you meant, and the other term is not standard. $\endgroup$
    – coffeemath
    May 1 '19 at 16:19
  • 1
    $\begingroup$ For another example of a distributive algebra that isn't power-associative see this question. $\endgroup$
    – pregunton
    May 1 '19 at 16:46

It seems to me you can construct an example by doing something like a group ring but not with a group, instead with a non-power-associative magma.

Let $M$ be any magma that isn't power-associative. You can just use the free magma on one element, whose elements start out $\{x, xx, x(xx), (xx)x, \ldots\}$.

Then, form formal combinations using integers as coefficients $\sum_{m\in M} z_mm$ which are stipulated to be finitely supported.

Multiplication is defined distributively, so the resulting algebra should be distributive, but it is also clearly not power-associative since $(xx)x\neq x(xx)$.


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