Defining groups by their inverse operation.

[Mark Hopkins]

A group can be defined by a single operation and 3 axioms, 2 when it's
Abelian. The axiomatizations are suprisingly simple.

-----------------

Consider the following axiom system:

SYSTEM A:
(1) A - (B - C) = C - (B - A)
(2) A - (A - B) = B

These properties hold for any Abelian Group, with the usual definition:

A - B = A + -B

An Abelian Group is recovered with the definitions:

-A = (A - A) - A and A + B = A - -B

In detail, from Axiom System A, we can prove the following:

(3) A - A = B - B for all A, B
Proof:
A - A = A - (B - (B - A)) by (2)
= A - (A - (B - B)) by (1)
= B - B by (2)

DEFINITION: Let 0 denote the unique element A - A. It follows that
-A = 0 - A, and A + B = A - (0 - B).

(4) A + -B = A - B
Proof:
A + -B = A - (0 - (0 - B))
= A - B by (2)

(5) A + B = B + A
Proof:
A + B = A - (0 - B)
= B - (0 - A) by (1)
= B + A

(6) A + 0 = A
Proof:
A + 0 = A - (0 - 0)
= A - (A - A) by (3)
= A by (2)

(7) A + -A = 0
Proof:
A + -A = A - (0 - (0 - A))
= A - A by (2)
= 0

(8) A - 0 = A
Proof:
A - 0 = A - (A - A) = A, by (3) and (2)

(9) (A + B) + C = A + (B + C)
Proof:
(A + B) + C = (A - (0 - B)) - (0 - C)
= C - (0 - (A - (0 - B))) by (1)
= C - ((0 - B) - (A - 0)) by (1)
= C - ((0 - B) - A) by (8)
= A - ((0 - B) - C) by (1)
= A - ((0 - B) - (C - 0)) by (8)
= A - (0 - (C - (0 - B))) by (1)
= A - (0 - (B - (0 - C))) by (1)
= A + (B + C)

Thus showing that Axiom System (A) with the indicated definitions describes
an Abelian Group.

THEOREM:
An Abelian Group is characterized as an algebraic system satisfying the
following axioms:
A - (B - C) = C - (B - A)
A - (A - B) = B
such that:
0 = A - A,
-A = (A - A) - A,
A + B = A - -B,
and A + -B = A - B

If an algebra satisfying Axiom System A is called an A-system, then define
an A-homomorphism as a map f: G -> H between two A-systems G and H such that:

f(u - v) = f(u) - f(v)
Then
f(0) = f(u - u) = f(u) - f(u) = 0
f(-u) = f(0 - u) = f(0) - f(u) = 0 - f(u) = -f(u)
f(u + v) = f(u - -v) = f(u) - f(-v)
= f(u) - -f(v) = f(u) + f(v).

Therefore, f is a group homomorphism between G and H. Conversely, assume that
f: G -> H is a homomorphism between groups G and H. Then, we have:

f(u - v) = f(u + -v) = f(u) + f(-v) = f(u) + -f(v) = f(u) - f(v)

showing that f is also an A-homormophism. So the homomorphism types for the
two algebraic systems also match.

Now consider the following axiom system:

SYSTEM B:
(1) A/(A/A) = A
(2) (A/A)/(B/B) = B/B
(3) (A/C)/(B/C) = A/B

These properties hold in any group, with the usual definition:

A/B = A B'

where ()' is being used to denote the inverse. A Group is recovered in a
similar way with the definitions:

A' = (A/A)/A and A B = A/B'

(4) A/A = B/B for all A and B
Proof:
A/A = (A/A)/(A/A) by (3)
= ((A/A)/(B/B))/((A/A)/(B/B)) by (3)
= (B/B)/(B/B) by (2), used twice
= B/B by (3) (or (2))

DEFINITION: Let 1 denote the unique element A/A. It follows that A' = 1/A
and AB = A/B'.

In the axiom system consisting of just (1) and (3), properties (2) and (4)
will be equivalent as can be seen by the following derivation:

(A/A)/(B/B) = (B/B)/(B/B) by (4)
= B/B by (3) (or (2))

Therefore, we could have equivalently defined an algebra satisfying the axiom
system B as that consisting of the binary operation /, the constant 1, subject
to the following properties:

(1') A/1 = A
(2') A/A = 1
(3') (A/C)/(B/C) = A/B

The properties of a group, with the operations defined as above are easily
established as follows:

(5) A/(B/(C/D)) = (A/(D/C))/B
Proof:
The following identity holds:
(B/(C/D))/(D/C) = (B/(C/D))/((D/D)/(C/D)) by (3)
= B/(D/D) by (3)
= B/(B/B) by (4)
= B by (1)
Using it, we get:
A/(B/(C/D)) = (A/(D/C))/((B/(C/D))/(D/C)) by (3)
= (A/(D/C))/B using the identity.

(6) B/1 = B
Proof:
B/1 = B/(B/B) = B by (1) and (4)

(7) A'' = A
Proof:
A'' = (1/A)'
= 1/(1/A)
= (A/A)/(1/A) by (4)
= A/1 by (3)
= A by (6)

(8) A B' = A/B
Proof:
A B' = A/B'' = A/B by (7)

(9) A 1 = A = 1 A
Proof:
A 1 = A/1' = A/(1/1) = A/1 = A by (4), (6)
1 A = 1/A' = A'' = A by (7)

(10) A A' = 1 = A' A
Proof:
A A' = A/A'' = A/A = 1 by (4), (7)
A' A = A'/A' = 1 by (4)

(11) (AB)C = A(BC)
Proof:
(AB)C = (A/B')/C'
= (A/(1/B))/(1/C)
= A/((1/C)/(B/1))) by (5)
= A/((1/C)/B) by (6)
= A/(((1/C)/(1/C))/(B/(1/C))) by (3)
= A/(1/(B/(1/C))) by (4)
= A/(1/(B/C'))
= A/(1/(BC))
= A/(BC)'
= A(BC)

where the unlabelled steps all throughout are applications of definitions.

The discussion about corresponding homomorphism types proceeds virtually
unaltered. If a B-algebra is defined as one satisfying Axiom System B,
then a B-homomorphism can be defined as a map f: G -> H between B-algebras
G and H satisfying the property:

f(u/v) = f(u)/f(v)

This completely characterizes a group homomorphism with the definitions
of the group operations given above.

[Herman Rubin]

In article <5tlq4q$2...@omnifest.uwm.edu>,

Mark Hopkins <ma...@omnifest.uwm.edu> wrote:
> A group can be defined by a single operation and 3 axioms, 2 when it's
>Abelian. The axiomatizations are suprisingly simple.

...............

The following is in the literature.

An Abelian group can be defined by the single equation, due to Tarski:

c = a - (b - (c - (a - b))).

There is a single equation of length 7 which defines a group by a
single operation. It has been shown that no equation of length 5
will work.

I do not know if this is published, but a Boolean group can be defined
by a single equation of length 5.

These are not recent, and I do not have references.

--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558