# Does Distributivity Imply Power Associativity?

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.

• @AniruddhAgarwal Induction on the size of the set? If so, I did try that, but I'll try again. If not, induction on what? – RothX May 1 at 16:15
• 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 – gt6989b May 1 at 16:16
• Would you be able to argue something from definition of multiplication via addition, like $$a \cdot (a \cdot a) = a \cdot (a + a + \ldots + a) = a\cdot a + \ldots + a\cdot a = (a \cdot a) \cdot a?$$ – gt6989b May 1 at 16:18
• 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. – coffeemath May 1 at 16:19
• For another example of a distributive algebra that isn't power-associative see this question. – pregunton May 1 at 16:46

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 $$(xx)x\neq x(xx)$$.