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When we can define a binary operation $\cdot:M\times M\rightarrow M$ on an algebraic structure $(M,*)$ such that

$$a*(b*c)=(a\cdot b)*c$$

If $*$ is associative then $\cdot=*$ even if I'm not sure about the uniqueness (But In right-invertible associative structures this is provable)

If $*$ is right-invetible then $a\cdot b=(a*(b*c))\setminus c$ only if $a\cdot b$ doesn't depends on $c$

So my question is

$1$-There is condition weaker than associativity for $*$ that make us able to define $a\cdot b$?

I am mainly interested in non-associative, right invertible and/or selfdistributive algebraic structures.

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One example that comes to mind is where $M=\mathbb{N}$, $a*b=b^a$, and $a\cdot b$ is ordinary multiplication. This also will often work more generally in settings where general exponentiation is well-defined (e.g. ordinals, cardinals...) – Harry Altman May 18 '14 at 21:23
up vote 2 down vote accepted

Well, there's the obvious equivalent condition: $$ \forall \, a,b,c \in M, \; \exists \, x \in M \text{ such that } a * (b * c) = x * c $$ Then we can just define $a \cdot b = x$.

This implies the stronger $$ \forall \, a_1, a_2, \ldots, a_k \in M, \; \exists \, x \in M \text{ such that } a_1 * (a_2 * (a_3 * \cdots * (a_{k-1} * a_k))) = x * a_k $$

Or, using the usual meaning of applying a binary operation to sets, $$ \forall \, c \in M, \; \forall k \in \mathbb{N} \; : \; M \underbrace{* (M * (M * \cdots *(M *}_{k \text{ } *\text{s}} \{c\}))) \subset M * \{c\} $$ I realize you probably want a weak condition that looks easy to check and not a strong condition like this, but I think this generates some insight into the nature of those $M$ for which $\cdot$ can be defined. Essentially, your property is saying: applying $M$ to the left of an set twice will produce a strictly smaller set than applying $M$ to the left of a set once.

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Thanks, I'll think about it. Anyways here there is a strictly related question:… There is even a bounty. – MphLee May 20 '14 at 6:35
The second stronger cndition is very interesting btw. I already know condition one, but I though there was an even weaker condition (easier to check...yes). I made this question because I had no answers in the other question about iteration of left traslations. – MphLee May 20 '14 at 6:43

There are several possibilities. I just want to mention an obvious one. Take any algebra $(M,*)$ which is "2-step nilpotent", i.e., satisfies $a\ast (b\ast c)=0$ for all $a,b,c$. Then we can take the zero product $a\cdot b=0$ to obtain $$ a\ast (b\ast c)=0=(a\cdot b)\ast c. $$ If the algebra $M$ satisfies $a\ast b= b\ast a$, or $a\ast b=-b\ast a$, also $a\cdot b=a\ast b$ is a possibility. For example, $M$ could be a two-step nilpotent Lie algebra (this is stronger than associativity, but one can certainly find examples with, say, "semi-associative" algebras, i.e., satisfying $(a,b,c)=(b,a,c)$ for the associator).

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Nice but there is a mre general requirement for the existence of $\cdot$? I mean if we define a class of algebraic structures where $\cdot$ exists what they have in common? In my question I was too broad. I don't ask only if there are weaker conditions but what is the weaker. About semiassociativity can you explain how we can define $\cdot$? – MphLee May 19 '14 at 10:13

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