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I've been thinking about an example of some set with a binary operation which would satisfy all axioms of groups except for associativity. I'm new to Group Theory, so I would appreciate your knowledgeable insight.

My example:

$S={\mathbb{C} - \{0\}}$; $H = (S, \text^)$, where ^ is exponentiation. Can we consider this as a correct example? The identity under ^ appears to be 1. The inverse appears to be $\frac{2i\pi}{\ln(x)}$.

sequence
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    $1$ is not an identity for ^ because although $x^1=x$, $1^x=1\ne x$, i.e. it is a right identity but not a left identity. In fact ^ has no two-sided identities. – Mario Carneiro Sep 21 '15 at 05:14
  • Does this mean that even if H were associative, it would not be a group? – sequence Sep 21 '15 at 05:15
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    Associativity is used as part of the proof that a left identity is a right identity and a left inverse is a right inverse. Without associativity you have to check both. – Mario Carneiro Sep 21 '15 at 05:16
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    Hint: look up loops. – Mario Carneiro Sep 21 '15 at 05:17
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    Even if we assume the identity element is two-sided, it is not certain that inverses would be two-sided as well. A normalized Latin square affords an example of a finite loop (nonassociative "group" operation). – hardmath Sep 21 '15 at 05:33
  • @hardmath Actually a Latin square just gives you a quasigroup - to that you also need to check for an identity, which corresponds to the top and left columns matching the table headers in a Cayley table, as in my example. – Mario Carneiro Sep 21 '15 at 05:57
  • @MarioCarneiro: By normalized Latin square I mean just that, the first row and column permuted to ascending order, i.e. matching the row and column headings. – hardmath Sep 21 '15 at 06:03
  • @hardmath I see, I was thinking that meant some sort of rescaling or something which didn't make any sense. Of course, if you start too small there is the danger that your loop ends up accidentally being a group - I believe the example in my answer to be minimal, as the same source mentions a result by R.P. Burn along the lines of "a loop of size $2p$ or $p^2$ is a group". – Mario Carneiro Sep 21 '15 at 06:07
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    Exponentiation is not even well-defined on the nonzero complex numbers. – Qiaochu Yuan Sep 21 '15 at 06:59
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    @QiaochuYuan It can be. It won't be very nice (continuous, say), but it's not like he's looking for a well-behaved operation with many identities, just the opposite. – Mario Carneiro Sep 21 '15 at 07:04

2 Answers2

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The object you are describing is called a loop. If you want a high powered example that fits in one mouthful, try the multiplicative group loop of non-zero octonions. As a more elementary alternative, I thought I'd give you a finite loop to ponder on (although you may need to work a bit to verify all the properties). I present Loop 8.1.4.0:

$$\begin{array}{c|c|c|c|c|c|c|c|} \circ& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7\\\hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7\\\hline 1 & 1 & 7 & 5 & 0 & 6 & 2 & 4 & 3\\\hline 2 & 2 & 6 & 7 & 5 & 0 & 3 & 1 & 4\\\hline 3 & 3 & 0 & 6 & 7 & 5 & 4 & 2 & 1\\\hline 4 & 4 & 5 & 0 & 6 & 7 & 1 & 3 & 2\\\hline 5 & 5 & 2 & 3 & 4 & 1 & 7 & 0 & 6\\\hline 6 & 6 & 4 & 1 & 2 & 3 & 0 & 7 & 5\\\hline 7 & 7 & 3 & 4 & 1 & 2 & 6 & 5 & 0\\\hline \end{array}$$

Here $0$ is the identity, and the inverses are $0,3,4,1,2,6,5,7$ for $0\dots7$ respectively. Consider $1\circ 1\circ 2$ to show non-associativity and $1\circ 2$ for non-commutativity.

The above is actually an example of a Bol loop, which satisfies the more complicated weak associativity property $a(b(ac))=(a(ba))c$. For general loops, there are smaller examples; the smallest non-associative loop has order $5$ - here is one of them:

$$\begin{array}{c|c|c|c|c|} \circ & 0 & 1 & 2 & 3 & 4\\\hline 0 & 0 & 1 & 2 & 3 & 4\\\hline 1 & 1 & 4 & 0 & 2 & 3\\\hline 2 & 2 & 3 & 4 & 1 & 0\\\hline 3 & 3 & 0 & 1 & 4 & 2\\\hline 4 & 4 & 2 & 3 & 0 & 1\\\hline \end{array}$$

Note that this does not have two-sided inverses, since we have identities like $1\circ 2=0$ and $3\circ 1=0$, so that the left inverse of $1$ is $2$ and the right inverse is $3$. For non-associativity just consider $1\circ 1\circ 1$ (this loop is not even power-associative).

MattAllegro
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Mario Carneiro
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  • For the Bol loop at the bottom, since left and right inverses are generally different, this would not be a group even if it were associative. For Loop 8.1.4.0, all properties (except for associativity) appears to be satisfied. – sequence Sep 21 '15 at 07:09
  • But I'm wondering if there's a set with a binary operation satisfying all group axioms except for associativity, which could be written in a short set notation? – sequence Sep 21 '15 at 07:11
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    @sequence Note that it doesn't make much sense to talk about "this would not be a group even if it were associative" for a particular example, because either it is associative or it isn't - there is no such thing as "if were associative", which is an explicitly contradictory hypothesis. If you consider adding the associativity axiom to the others, in general, then the left/right inverses thing would disappear at the same time - an associative loop is a group, i.e. there are no loops with associativity but not two-sided inverses. – Mario Carneiro Sep 21 '15 at 07:18
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    @sequence As for "written in a short set notation", the main thrust of the first paragraph is that most examples are either non-elementary or complicated. The octonion example can be written easily as $\Bbb O\setminus{0}$ but of course that hides a lot of detail - it's a non-elementary example. On the other hand you have finite loops, which need no advanced definitions but are hard to present concisely. – Mario Carneiro Sep 21 '15 at 07:21
  • That means all latin squares of order $<5$ are "at least" semi groups? – Raphael J.F. Berger Feb 14 '18 at 16:50
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Consider the set $\{0,1,2\}$ with the binary operation $a*b = |a-b|$. It has an identity element (namely $0$) and every element has an inverse (namely itself) and it is not associative: $(1*1)*2 = 0*2 = 2$ but $1*(1*2) = 1*1 = 0$.

James Propp
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  • Hmm. That's interesting. This solution appears to answer the question which was asked, but it also contradicts the accepted answer. This is because the accepted answer said that the questioner was looking for a "Loop", while James Propp's answer does not describe a Loop. You can see this because every loop is a quasigroup, and every quasigroup should have unique solutions to ax=b problems. Yet 1*x=1 does not have a unique solution here: x could be both 0 or 2. – KesterKester Feb 26 '21 at 00:30
  • @KesterKester I would argue that the OP was ambiguous about what it means to "satisfy all axioms of groups except for associativity" or whether "has inverses" implies uniqueness or not. Depending on the resolution you could either interpret the result as in James's answer or mine. – Mario Carneiro Sep 12 '23 at 00:41
  • That is a fair comment. There are a number of different ways that "a group" could be defined as a structure containing "Associativity" in addition to some other property or properties which we might call "X". While any valid X plus assciativity would lead to a group, it is not true that all X's without asscoativity are equivalent. So, yes, I agree with @MarioCarneiro's point. E.g. two inequivalent possibilities for X could be (1) "is disvisible and non-emtpy", and (2) "is divisible and has a unique inverse". Either of those plus associativity is a group, but X1 is not the same as X2. – KesterKester Sep 28 '23 at 14:25