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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?

Danae Kissinger
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  • @Alizter I think that plus is commutative, and minus certainly not, but I can't see why this should be a problem – Danae Kissinger Apr 08 '15 at 13:14
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    @Alizter This is precisely what I am asking: if I use minus in place of plus, what will be new axioms describing the structure of abelian groups? What you are saying to me is simply: commutativity won't appear in the new set of axioms. – Danae Kissinger Apr 08 '15 at 13:17
  • @alitzer because he wants an abelian group which described in terms of the$-$ operation. You can take a look at the answer to see what the question was asking. – DRF Apr 08 '15 at 13:35
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    Related reference: http://www.jstor.org/discover/10.2307/2307287?sid=21106391248673&uid=3738296&uid=2&uid=4 – MattAllegro Apr 08 '15 at 17:10

3 Answers3

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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.

Andreas Caranti
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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.

Martin Brandenburg
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  • Technically, you don't even need zero, just an axiom $x-x=y-y$ and then use $x-x$ in all your axioms where you used $0$. – Thomas Andrews Apr 08 '15 at 22:20
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    Not quite, since this would include $X=\emptyset$ (the same is true for Andreas Caranti's answer). And the axiom $X \neq \emptyset$ does not fit to universal algebra. Also, as explained in my answer, I didn't want to have some minimal system of axioms. It should be natural instead. – Martin Brandenburg Apr 08 '15 at 22:32
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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.

Najib Idrissi
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