(I give a lengthy introduction to a concise question -- scroll down if you want to jump straight up to the question).

Recall that abelian group theory consists of two primitive symbols: $\cdot$ which is a binary function symbol, and $e$ which is a constant. The axioms are:

($G_1$) $\forall x \forall y \forall z \ \ x\cdot (y\cdot z)=(x\cdot y)\cdot z$

($G_2$) $\forall x \ \ x\cdot e=x$

($G_3$) $\forall x \exists y \ \ x\cdot y=e$

($G_4$) $\forall x \forall y \ \ x\cdot y =y \cdot x$

A set of axioms $\Phi$ is independent if for every $\varphi\in \Phi$ there exists an interpretation $\mathcal{I}$ such that $\mathcal{I} \models \Phi\setminus \{\varphi\}$ and $\mathcal{I}\not\models \varphi$.

In non-formal logic language, $\mathcal{I}\models \Phi$ means: exhibit a set $G$ with a binary operation $\cdot$ and an element $e\in G$ such that all axioms in $\Phi$ are satisfied (taking the variables as belonging to $G$).

So, to prove that the above axioms are independent is to exhibit, for every $i=1,\dots,4$, a set $G$ with a binary operation $\cdot$ and an element $e\in G$ such that $(G_j)$ holds for every $j\not=i$, and $(G_i)$ does not hold. Thusly you prove that you can't prove $(G_i)$ from $\{(G_j), j\not=i\}$.

A cute and fun problem in Ebbinghaus, Mathematical Logic (exercise 4.14, p. 39) asks us to prove that the group theory axioms, i.e. $\{(G_1), (G_2), (G_3)\}$ is an independent set of axioms. This is fun to do.

But then the natural follow-up question that occurred to me is: is $\{(G_1), (G_2), (G_3) ,(G_4)\}$ an independent set of axioms?

For $i=2,3,4$ it is easy to prove that there are models of $\{(G_j):j\not=i\}$ where $(G_i)$ does not hold. (In fact, for $i=2,3$, the ones I have thought for the exercise in Ebbinghaus were all commutative, and thus worked; for $i=4$ it's just the existence of non-abelian groups).

But for $i=1$ I'm having a really hard time. I tried a lot of examples, neither of which works. To sum up, I'm trying to prove that:

There exists a set $G$ with a binary operation $\cdot$ such that: $\cdot$ is not associative, $\cdot$ is commutative, there is an identity element $e$, and every element has an inverse with respect to $e$.

The best I could do was the following. Take $G=\mathbb{R}^2$, with $(a,b)\cdot (c,d)=(ac+bd,0)$. It is commutative, not associative, the inverse with respect to $(0,0)$ is $(b,-a)$, but $(a,b)\cdot (0,0)=(0,0)$, not $(a,b)$, whence $(0,0)$ is not an identity element.

Or perhaps I'm just wrong and $(G_2), (G_3), (G_4)$ imply $(G_1)$, which would come off as a surprise.

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    $\begingroup$ You are looking for a commutative loop that is not a group. $\endgroup$ Commented Aug 6, 2011 at 22:02

3 Answers 3


Consider the non-negative integers $\mathbb{Z}_{\ge 0}$ with the binary operation $|x - y|$. This is obviously commutative. The identity is $0$ and every element is its own inverse, but $||1 - 1| - 2| = 2 \neq |1 - |1 - 2|| = 0$.

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    $\begingroup$ Fantastically simple, thanks! Makes me feel a little dumb though for not finding an example after playing around so much :) $\endgroup$ Commented Aug 6, 2011 at 22:14

Consider the set with three elements $\{e, x, y\}$. Define $\cdot$ to be $a \cdot e = a$ for all $a$, $e \cdot a = a$ for all $a$, and $a \cdot b = e$ for all $a$ and $b$ such that $a \neq e$ and $b \neq e$.

Then clearly this operation is commutative, had identity $e$, has inverses, but $(x \cdot y) \cdot y = e \cdot y = y$ and $x \cdot (y \cdot y) = x \cdot e = x$.

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    $\begingroup$ Nice. I do find this somewhat of being a smartass, as the inverse is usually unique in a group. In this case, however, it is not. Nice way of exploiting this fact! $\endgroup$
    – Asaf Karagila
    Commented Aug 6, 2011 at 22:13
  • $\begingroup$ Very nice! I was thinking of describing the free example (which is fairly complicated), but this is something like the "cofree" example and it is quite simple. $\endgroup$ Commented Aug 6, 2011 at 22:16
  • $\begingroup$ Thanks, this is a very nice example too, especially taking into account Asaf's comment. @Qiaochu: care to elaborate? $\endgroup$ Commented Aug 6, 2011 at 22:19
  • $\begingroup$ @Bruno: there should be a free widget on two generators with a binary operation which is commutative and has an identity and inverses, and this thing shouldn't be associative. It can be described as a collection of trees up to a fairly complicated equivalence relation, so it's not easy to prove things about it. But William's idea to set everything that wasn't constrained to be something else to the identity is much neater. $\endgroup$ Commented Aug 6, 2011 at 22:29

Here's one that's commutative, has an identity (0), each element has a unique inverse, but it's not associative: $$\matrix{0&1&2&3&4&5\cr1&0&3&4&5&2\cr2&3&0&5&1&4\cr3&4&5&0&2&1\cr4&5&1&2&0&3\cr5&2&4&1&3&0\cr}$$ If it were associative, it would be a group of order 6 in which each element has order 2, but there's no such group, so it can't be associative.


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