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I don't know if this question doesn't answer itself.

Generally, when a binary operation $\circ:S\times{S}\to{S}$ is called associative, it is usually sufficient to state:

$$a\circ(b\circ{c})=(a\circ{b})\circ{c}$$

defines the operation as associative.

I don't know whether, if the above identity holds, it can be proven by induction that all parenthesizations of an expression containing are equivalent. I therefore ask the following in some ignorance:

Are there algebraic objects in which "associativity" holds only up to some finite $n$?

That is:

Are there algebraic objects in which, e.g, $a\circ(b\circ{c})=(a\circ{b})\circ{c}$ holds but $a\circ({b}(\circ{c}(\circ{d})))=(((a\circ{b})\circ{c})\circ{d}$ does not?

Michael McGovern
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1 Answers1

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Yes, by induction associativity implies higher-arity associativity.

For example, if I want to show that $a\circ (b\circ (c\circ d))=((a\circ b)\circ c)\circ d,$ I can argue as follows:

  • $a\circ (b\circ (c\circ d))=(a\circ b)\circ (c\circ d)$.

  • $(a\circ b)\circ (c\circ d)=((a\circ b)\circ c)\circ d$.

The full proof by induction is a good exercise.

There are, however, important weak forms of associativity coming from restrictions on the elements, e.g. power associativity and alternativity. Note that for weaker versions of associativity, the "full" version may not follow from the "basic" version, e.g. in the case of power associativity it's not enough to just demand $x\circ (x\circ x)=(x\circ x)\circ x$ for all $x$.

Noah Schweber
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