Is there a name for the generalization of the concept "Abelian group" where the axiom $-x+x = 0$ is weakened to the following?

Question

Is there a name for the generalization of the concept "Abelian group" where the axiom $−x+x=0$ is replaced by the following list?

$−0=0$
$−(x+y)=−x+−y$
$−(−x)=x$
$x+(-x)+x = x$

In multiplicative notation; we replace the axiom $x^{-1}x=1$ with the following list:

$1^{-1}=1$
$(xy)^{-1}=x^{-1}y^{-1}$
$(x^{-1})^{-1}=x$
$xx^{-1}x = x$

Examples.

Any Abelian group satisfies the above axioms in their additive form.
The multiplicative structure of any zero-totalized field satisfies the above axioms in their multiplicative form, but does not satisfy $x^{-1}x=1$, since $0^{-1} \cdot 0 = 0 \cdot 0 = 0$.

Concerning your notion of a "zero-totalized field": If $0^{-1} = 0$ in any field, then $00 = 1$. But this contradicts the basic property $a0 = 0a = 0$ which holds in any ring. — Ulrik, Mar 03 '14 at 11:37
@Svinepels, no; the reciprocation axiom in the language of fields is $\forall x(x \neq 0 \rightarrow xx^{-1}=1),$ and not $\forall x(xx^{-1}=1)$. — goblin GONE, Mar 03 '14 at 11:55
@TobiasKildetoft, do they really satisfy axiom 4? I don't think this is correct, since not every matrix is skew-symmetric. — goblin GONE, Mar 03 '14 at 11:57
@user18921: Ok, so if I understand correctly, $0^{-1} = 0$ doesn't imply $00 = 1$? — Ulrik, Mar 03 '14 at 12:02
@Svinepels yes, all we can deduce is that $0 \neq 0 \rightarrow 00^{-1}=1$, which is vacuous. — goblin GONE, Mar 03 '14 at 12:03
@TobiasKildetoft yep I stuffed up the first draft, not your fault. — goblin GONE, Mar 03 '14 at 12:04
@KevinCarlson, because the axiom $xx^{-1}x=x$ is equivalent to $xx^{-1}x = x1,$ so this basically says that in the presence of $x$, the expression $x^{-1}x$ behaves as expected. This is a familiar idea from linear algebra; see also Moore-Penrose pseudoinverse. — goblin GONE, Mar 06 '14 at 01:34
Following the link you gave me, it looks like this is just a commutative regular semigroup with identity on which you've chosen a pseudo-inverse for each element. — Kevin Carlson, Mar 06 '14 at 01:42
@KevinCarlson, hmm perhaps. Axiom 1 certainly follow from 4; if we can show that both 2 and 3 follow, then you are correct. — goblin GONE, Mar 06 '14 at 01:54

score 4 · Accepted Answer · answered Mar 06 '14 at 02:24

4

Apparently it's called a (commutative) inverse monoid. For further details, see Wikipedia^{1, 2, 3} or Lawson's Inverse Semigroups⁴.

(I haven't proven that the sets of axioms are equivalent. You may want to reserve the bounty for someone who does so.)

answered Mar 06 '14 at 02:24

Chris Culter

27,415

Is there a name for the generalization of the concept "Abelian group" where the axiom $-x+x = 0$ is weakened to the following?

1 Answers1