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I seem to recall that there is a relatively easy method for determining the associativity of an operation by using its Cayley table. What is it?

user1729
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Stephen Meskin
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  • I disagree that this question should give context or other detail - it admits a well-defined and interesting answer. The "context" is given: "can you help me remember something?" If the answer was not well defined then I can see an issue, but as it is well-defined... – user1729 Oct 24 '18 at 12:38
  • Please look through How to ask a good question for advice on writing good questions on this site. In particular, a post should go beyond merely stating a problem: the motivation and background should be included, to the extent you are familiar. Posts that merely state a problem without context are often put on hold. – Carl Mummert Oct 28 '18 at 01:06

1 Answers1

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It is called Light's associativity test which I found on Wikipedia.

Basically,

  1. Pick out the generators of the operation.
  2. If $g$ is a generator define two new operations $x \circ y = (xg)y $ and $x*y=x(gy)$.
  3. Form the Cayley tables of $\circ$ and $*$ for $g$.
  4. If the two tables for $g$ are not identical, the original operation is NOT associative.
  5. If the two tables are identical for all generators $g$, the original operation IS associative.

Notwithstanding the first comment, the link above works now, Nov 7, thanks to a kind editor.

Stephen Meskin
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  • The URL you give doesn't work. The method you describe doesn't save much work unless it's easy to find a set of generators that is much smaller than the number of elements in the structure. – Rob Arthan Nov 07 '17 at 22:39
  • @RobArthan You can always find a generating set of size at most $\log_2 n$ in $O(n^2)$ time for a group of $n$ elements described by a Cayley table. This takes less time than the test itself. – Qudit Nov 07 '17 at 23:06
  • @Qudit: that's interesting, but I was thinking more of hand calculation for small magmas (say with less 10 elements) rather than the asymptotic complexity of the method. It's easy to come up with an $n$-element magma that has no generating set with less than $n$ elements, so I don't see how your observation about generating sets of groups helps in either case (if you know the magma is a group, then you already know it's associative.). – Rob Arthan Nov 07 '17 at 23:29
  • @RobArthan That's true. I suppose that in the case where you were testing for associativity in order to check if it is a group, you could try to find a generating set of size at most $\log_2 n$. If the procedure fails to find one in $O(n^2)$ time, then you know that it is not a group so you don't care if it is associative. Otherwise, you can proceed with the test. – Qudit Nov 07 '17 at 23:38
  • @Qudit: but the question (and Light's test) is about testing for associativity, not testing for all the group properties: people often do care about the associativity of operations that are not group operations. In any case, I think you need to modify your update to the Wikipedia page. – Rob Arthan Nov 07 '17 at 23:46
  • @RobArthan Yes, I already reverted it after you posted your initial reply. – Qudit Nov 07 '17 at 23:52
  • @Qudit: I see you have undone your update to the Wikipedia page: thanks for that. Perhaps your implied conjecture about the complexity of determining whether a finite magma is a group is worth posing as an MSE question. I think you are conjecturing that it is no worse than $O(n^2 \log(n))$. Can one do better than that? Hmmm! – Rob Arthan Nov 07 '17 at 23:53
  • @RobArthan Actually, there seems to be an easier way to prove that it can be done in $O(n^2 \log n)$ time. First, check if the magma is a quasigroup in $O(n^2)$ time. Quasigroups also have generating sets of size at most $\log n$, so you can then apply Light's associativity test. I am not sure if you can get rid of the $\log n$ deterministically though. Wikipedia claims that there is a $O(n^2)$ BPP algorithm at the end of the article. – Qudit Nov 08 '17 at 00:04
  • @RobArthan See https://math.stackexchange.com/questions/2509960/complexity-of-testing-if-a-binary-operation-is-a-group – Qudit Nov 08 '17 at 00:25
  • The purpose of the OP was to help people learning algebra who are having a hard time imagining doing the $216$ calculations they would have to do to check the associative law on a set of only three elements. I'm glad that it has also spurred a fascinating conversation between Rob Arthan and Qudit on a much deeper subject. – Stephen Meskin Nov 08 '17 at 01:43