56
$\begingroup$

Most properties of a single binary operation can be easily read of from the operation's table. For example, given $$\begin{array}{c|ccccc} \cdot & a & b & c & d & e\\\hline a & e & d & b & a & c\\ b & d & c & e & b & a\\ c & b & e & a & c & d\\ d & a & b & c & d & e\\ e & c & a & d & e & b \end{array}$$ it is easy to check that it is closed (no elements occur in the table which don't occur as row or column index), commutative (the table is symmetric), has a neutral element (the row and column of $d$ are copies of the index row/column), and has an inverse element for each element (there's a $d$ in each row and column). In other words, almost all important properties can immediately be seen. The only part missing is associativity.

Therefore my question: Is there a simple way to see directly from the operation's table (i.e. without doing explicitly all the calculations) if an operation is associative?

$\endgroup$
0
31
$\begingroup$

Have you seen Light's associativity test? According to Wikipedia, "Direct verification of the associativity of a binary operation specified by a Cayley table is cumbersome and tedious. Light's associativity test greatly simplifies the task."

If nothing else, the existence of Light's algorithm seems to rule out the possibility that anyone knows an easy way to do it just by looking at the original Cayley table.

Note also that, in general, one cannot do better than the obvious method of just checking all $n^3$ identities of the form $(a\ast b)\ast c = a\ast (b\ast c)$. This is because it is possible that the operation could be completely associative except for one bad triple $\langle a,b,c\rangle$. So any method that purports to do better than this must only be able to do so in limited circumstances.

$\endgroup$
1
  • 6
    $\begingroup$ This is not a valid proof that checking $n^3$ identities individually is the only valid algorithm. There may be an algorithm that checks multiple identities at once by using the Cayley table in some smart way. $\endgroup$
    – exfret
    Nov 20 '18 at 21:34
18
$\begingroup$

Using the original $n\times n$ table seems bleak - this is essentially a problem of arity-dimension three, but the Cayley table only gives us two dimensions. However, Light's Associativity Test shows how to systematically reduce the problem of comparing $n$ pairs of Cayley tables. Note that the procedure can be greatly simplified by considering only operations derived from the underlying structure's generators.

$\endgroup$
1
  • $\begingroup$ Thank you (and also the others). This test works in a mechanical way (copying rows/columns around and comparing tables), so it's alreafy quite close to actually "seeing" the result (one might argue that the only reason why one cannot directly see the result of this test is that the short-time memory of the human brain has not enough capacity). I've chosen to accept your answer because it was slightly earlier than Mark's (which would also have qualified as the accepted answer, but I had to choose one). $\endgroup$
    – celtschk
    Jul 9 '12 at 16:40
11
$\begingroup$

First of all, let me make a personal reflection on this matter. Light's associativity test (as others have noted) provides a characterization, but (at least from my point of view) it is not really helpful. Indeed, I like to consider this difficulty to check whether a table is associative as the main reason why it is better to introduce associative operations (in particular groups) through presentations. Then, you trivially get associativity since your "object" is by definition a quotient of the free one.

Now, let me note that in the particular case that the operation is commutative (like the example you have written) then it is known an alternative method which is affordable to be done using a pencil. This (quite unknown in my opinion) method is due to S. KAMAL ABDALI and was introduced in his paper "Verification of Associativity of a Binary Operation" http://www.jstor.org/stable/3613856

I have never seen this method explained in a book, so it is worthwhile to take a look at this paper (in case you can go through the publisher firewall).

$\endgroup$
2
  • $\begingroup$ This only seems to reduce the work needed by a constant factor. How is this really any better than just checking each triple individually? $\endgroup$
    – exfret
    Nov 20 '18 at 21:44
  • $\begingroup$ Reducing the work by a factor is not nothing, but the (computational) order is still $n^3$; you have to show $(xy)z = (yz)x = (zx)y$ and that triplet is done, but the choice of triplets is $\binom n3$ which has an $n^3$ term.. $\endgroup$
    – Joffan
    Apr 18 at 20:33
9
$\begingroup$

There is a randomized algorithm solving this problem in time proportional to the input size. Specifically, the runtime is $O\left(n^2 \log\frac1\delta\right)$ for an $n \times n$ table and error probability $\delta$. See Rajagopalan and Schulman, FOCS 1996, SICOMP 2000. http://dx.doi.org/10.1137/S0097539797325387

Free download here.

$\endgroup$
2
  • $\begingroup$ I read and enjoyed this paper. Thanks very much for the reference. $\endgroup$
    – MJD
    Oct 18 '13 at 2:55
  • $\begingroup$ lecture note $\endgroup$
    – miracle173
    Jul 27 '16 at 21:29
2
$\begingroup$

If I remember correctly, this is explained in W. Keith Nicholson's Introduction to Abstract Algebra. I don't have a copy with me in my office, but I will double-check when I get home.

Edit: I believe that what is being described in the book cited above is exactly Light's Associativity Test, which is mentioned in the other answers.

$\endgroup$
1
$\begingroup$

In R.P. Burn's article, Cayley Tables and Associativity (paywall but JSTOR a lot easier to access than most), there are provided some interesting ways to "read off" weaker forms of associativity (such as $a(ax)=(aa)x$) from Cayley tables.

Presumably trying to combine them all to get full associativity would be just as bad as any other algorithm, but in certain circumstances it could, perhaps, suffice. At any rate it is something from the original Cayley table and not derived ones.

$\endgroup$
-3
$\begingroup$

A set with an operation used to build a four-dimensional multiplication table with entries $X_{abcd} = (A(BC))D$ is associativ iff $X_{abc1} = X_{1abc}$, with 1 being unity. So it would be a certain three-dimensional symmetry of a four-dimensional table.

$\endgroup$
1
  • 2
    $\begingroup$ In this notation the above cited test for commutative operations from S. Kamal Abdali cheks the symmetry of the 3D table: $X_{abc} = A(BC)$. It is checked if $X_{abc} = X_{cba}$, and just using only a lower diagonal due to commutative symmetry. For non-commutative operations, one needs to go to the above stated symmetry of a 4D table. $\endgroup$ Nov 5 '15 at 21:29

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.