1
$\begingroup$

With the given information, I'm trying to prove that an operation is both associative and commutative.

Let $(A,*)$ be a set and a binary operation such that for all $a,b,c \in A$ $$a*(b*c) = a*(c*b),$$ and for some $e \in A$ and for any $a \in A$ $$e*a = a*e = a.$$

So far, proving that $*$ is commutative has been easy, but finding a method to prove associativity has escaped me. I don't want the solution to the problem, but could anyone help in assuring me that associativity is even possible with the given assumptions.

Thanks!

$\endgroup$
2
$\begingroup$

No.

Counterexample:

$A = \{a,b,c \}$

\begin{array}{c|ccc} *&a&b&c\\ \hline a&a&b&c\\ b&b&c&c\\ c&c&c&b\\ \end{array}

$*$ is clearly commutative (and hence also $a * (b * c) = a * (c * b)$), $a$ is its identity element, but $*$ is not associative:

$b*(b*c)=b*c=c$ but $(b*b)*c=c*c=b$

$\endgroup$
2
$\begingroup$

It is not possible. Consider the operation of combining unordered binary trees, where a * b is the binary tree whose root's unordered pair of children are the roots of a and b; toss in by fiat also an identity element for this operation. This operation is clearly commutative and thus satisfies your properties, but does not satisfy associativity.

For a less abstract example, consider something like the operation $|a - b|$.

$\endgroup$

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.