Are the axioms for abelian group theory independent?

Question

(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.

Answer 1

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 $||1 - 1| - 2| = 2 \neq |1 - |1 - 2|| = 0$.

Answer 2

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 $x \cdot (y \cdot y) = x \cdot e = x$.

Answer 3

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.