Abelian groups axioms with minus in place of plus An abelian group is a set equipped with a binary operation $+$, a unary operation $-$ and a nullary operation (constant) $0$, satisfying certain axioms (associativity, unity, etc).
I wonder if it is possible to describe the same structure, that is that of abelian group, using a binary operation $-$ and a constant $0$. I think it is possible by setting $a-b=a+(-b)$ where in the LHS there is the binary $-$ I am defining, while in the RHS we have the usually binary $+$ and the unary "inverse" operator $-$. But which should be the axioms for $-$ in order to have an abelian group?
 A: Once you know that you can rewrite $a + b = a - (0 - b)$, it's simply a matter of going through the abelian group axioms and rewriting everything:

An abelian group is a triplet $(A, -, 0)$ consisting of a set $A$, a law $- : A \times A \to A$ and an element $0 \in A$ such that:
  
  
*
  
*Associativity: for all $a,b,c \in A$,
  $$(a + b) + c = a + (b + c) \iff \color{red}{(a - (0 - b)) - (0 - c) = a - (0 - (b - (0 - c)))}$$
  
*Identity element: for all $a \in A$,
  $$a + 0 = a = 0 + a \iff \color{red}{a - (0 - 0) = a = 0 - (0 - a)}$$
  
*Inverse element: for all $a \in A$, there exists $b \in A$ such that:
  $$a + b = 0 = b + a \iff \color{red}{a - (0 - b) = 0 = b - (0 - a)}$$
  
*Commutativity: for all $a, b \in A$,
  $$a + b = b + a \iff \color{red}{a - (0 - b) = b - (0 - a)}$$

To get a "true" abelian group from this kind of abelian group, simply set $a + b = a - (0 - b)$, and you'll see that all the axioms are there.
A: There's Tarski's famous single axiom for Abelian groups in terms of subtraction:
$$
x - (y - (z - (x - y))) = z.
$$

See here for a version by G. Higman and B. H. Neumann for groups.
A: Let $(X,-,0)$ be a triple consisting of a set $X$, a binary operation $-$ on $X$ and a constant $0$ in $X$. We will require the following axioms:


*

*$(a-d)-(b-c)=(a-b)-(d-c)$

*$a-0 = a$

*$a-a=0$

*$b-(c-a)=a-(c-b)$


Define $a+b := a-(0-b)$. I claim that $(X,+,0)$ is an abelian group.
It is clear that $a+b=b+a$, $a+0=a$, and $(0-a)+a=0$. Finally, $+$ is assocative: We start with
$$(a-0)-(b-c)=(a-b)-(0-c)$$
Substituting $b$ by $0-b$ and using $a-0=a$ yields
$$a-((0-b)-c)=(a-(0-b))-(0-c)$$
The right hand side is clearly $(a+b)+c$. The left hand side equals $a+(b+c)$ because
$$0-(b+c) = 0 -(b-(0-c)) = (0-0)-(b-(0-c))\\
=(0-b)-(0-(0-c)) =(0-b)-(c-(0-0))=(0-b)-c.$$
Thus, $(X,+,0)$ is an abelian group.
Conversely, if $(X,+,0)$ is an abelian group, then $a-b := a+(-b)$ satisfies the axioms. These constructions are inverse to each other. The notions of homomorphisms also agree, so that we get isomorphic categories.
Notice that it is possible to reduce the axioms and still make the proof work: In axiom 1, we may assume $d=0$. In axiom 4, we may assume $c=0$. But I think that the axioms look more natural when stated as above.
