more on division in groups

more on division in groups

In the parent entry, it is shown that a non-empty set

G

equipped a

binary operation

“

/

” called “division” satisfying three

identities

has the

structure

of a group. In this entry, we show that two identities are enough. Associated with every

x

,

y

∈

G

, we set

1.

E

⁢

(

x

)

:=

x

/

x

,

2.

(

inverse

)

x

-

1

:=

E

⁢

(

x

)

/

x

, and

3.

(

multiplication

)

x

⋅

y

:=

x

/

y

-

1

(we also write

x

⁢

y

for

x

⋅

y

for simplicity)

Theorem 1

.

Let

G

be a non-empty set with a binary operation

/

on it such that

1.

(

x

/

z

)

/

(

y

/

z

)

=

x

/

y

2.

(

x

/

x

)

/

(

(

y

/

y

)

/

y

)

=

y

hold for all

x

,

y

,

z

∈

G

. Then

G

has the structure of a group

Proof.

From 1, we have

E

⁢

(

x

/

z

)

=

(

x

/

z

)

/

(

x

/

z

)

=

x

/

x

=

E

⁢

(

x

)

, so

E

⁢

(

E

⁢

(

x

)

)

=

E

⁢

(

x

/

x

)

=

E

⁢

(

x

)

. From 2, we have

y

=

E

⁢

(

x

)

/

y

-

1

, so

E

⁢

(

y

)

=

E

⁢

(

E

⁢

(

x

)

/

y

-

1

)

=

E

⁢

(

E

⁢

(

x

)

)

=

E

⁢

(

x

)

. This shows that

E

:

G

→

G

is a

constant function

, whose value we denote by

e

.

Note that

x

=

e

/

x

-

1

by rewriting condition 2. This implies that

e

⋅

x

=

e

/

x

-

1

=

x

. In

addition

,

x

-

1

=

e

/

x

by rewriting the definition of the inverse. In particular,

e

-

1

=

e

/

e

=

e

. Furthermore, since

x

/

e

=

(

e

/

x

-

1

)

/

(

x

-

1

/

x

-

1

)

=

e

/

x

-

1

=

x

, this implies that

x

⋅

e

=

x

/

e

-

1

=

x

/

e

=

x

. So

e

is the “identity” in

G

with respect to

⋅

.

Next,

x

-

1

⋅

x

=

x

-

1

/

x

-

1

=

e

. To see that

x

⋅

x

-

1

=

e

, first observe that

(

x

-

1

)

-

1

=

e

/

x

-

1

=

x

, so

x

⋅

x

-

1

=

x

/

(

x

-

1

)

-

1

=

x

/

e

=

x

. This shows that

x

-

1

is the “inverse” of

x

in

G

with respect to

⋅

.

Finally, we need to verify

(

x

⁢

y

)

⁢

z

=

x

⁢

(

y

⁢

z

)

. To see this, first note that

1.

(

x

⁢

y

)

/

y

=

(

x

/

y

-

1

)

/

y

=

(

x

/

y

-

1

)

/

(

e

/

y

-

1

)

=

x

/

e

=

x

, and

2.

(

x

⁢

y

)

-

1

=

e

/

(

x

⁢

y

)

=

e

/

(

x

/

y

-

1

)

=

(

y

-

1

/

y

-

1

)

/

(

x

/

y

-

1

)

=

y

-

1

/

x

=

y

-

1

⁢

x

-

1

.

From the two identities above, we deduce

(

x

⁢

y

)

⁢

z

=

(

x

⁢

y

)

/

z

-

1

=

(

x

/

y

-

1

)

/

z

-

1

=

(

x

/

y

-

1

)

/

(

(

z

-

1

⁢

y

-

1

)

/

y

-

1

)

=

x

/

(

z

-

1

⁢

y

-

1

)

=

x

/

(

y

⁢

z

)

-

1

=

x

⁢

(

y

⁢

z

)

,

completing the proof.
∎

There is also a companion

theorem

for

abelian groups

:

Theorem 2

.

Let

G

be a non-empty set with a binary operation

/

on it such that

1.

x

/

(

x

/

y

)

=

y

2.

(

x

/

y

)

/

z

=

(

x

/

z

)

/

y

hold for all

x

,

y

,

z

∈

G

. Then

G

has the structure of an abelian group

Proof.

First, note that

E

⁢

(

x

/

y

)

=

(

x

/

y

)

/

(

x

/

y

)

=

(

x

/

(

x

/

y

)

)

/

y

=

y

/

y

=

E

⁢

(

y

)

, so

E

⁢

(

x

)

=

E

⁢

(

(

x

/

y

)

/

x

)

=

E

⁢

(

(

x

/

x

)

/

y

)

=

E

⁢

(

y

)

, implying that

E

is a constant function on

G

. Again, denote its value by

e

. Below are some simple

consequences

:

1.

x

/

e

=

x

/

(

x

/

x

)

=

x

2.

e

-

1

=

e

/

e

=

e

3.

(

x

-

1

)

-

1

=

(

e

/

x

)

-

1

=

e

/

(

e

/

x

)

=

x

So,

x

⁢

e

=

x

/

e

-

1

=

x

/

e

=

x

. Also,

e

⁢

x

=

e

/

x

-

1

=

e

/

(

e

/

x

)

=

x

. This shows that

e

is the “identity” of

G

with respect to

⋅

. In addition,

x

-

1

⁢

x

=

x

-

1

/

x

-

1

=

e

and

x

⁢

x

-

1

=

x

/

(

x

-

1

)

-

1

=

x

/

x

=

e

, showing that

x

-

1

is the “inverse” of

x

in

G

with respect to

⋅

.

Finally, we show that

⋅

is

commutative

and

associative

. For commutativity, we have

x

⁢

y

=

(

e

⁢

x

)

⁢

y

=

(

e

/

x

-

1

)

/

y

-

1

=

(

e

/

y

-

1

)

/

x

-

1

=

(

e

⁢

y

)

⁢

x

=

y

⁢

x

, and associativity is shown by

x

⁢

(

y

⁢

z

)

=

(

y

⁢

z

)

⁢

x

=

(

y

/

z

-

1

)

/

x

-

1

=

(

y

/

x

-

1

)

/

z

-

1

=

(

y

⁢

x

)

⁢

z

=

(

x

⁢

y

)

⁢

z

.
∎

Remark

. Remarkably, it can be shown (see reference) that a non-empty set

G

with binary operation

/

satisfying a single identity:

x

/

(

(

(

(

x

/

x

)

/

y

)

/

z

)

/

(

(

(

x

/

x

)

/

x

)

/

z

)

)

=

y

has the structure of a group, and satisfying

x

/

(

(

y

/

z

)

/

(

y

/

x

)

)

=

z

has the structure of an abelian group.

References

1

G. Higman, B. H. Neumann

Groups as

groupoids

with one law

. Publ. Math. Debrecen 2 pp. 215-221, (1952).

Title

more on

division in groups

Canonical name

MoreOnDivisionInGroups

Date of creation

2013-03-22 17:38:22

Last modified on

2013-03-22 17:38:22

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

8

Author

CWoo (3771)

Entry type

Result

Classification

msc 08A99

Classification

msc 20A05

Classification

msc 20-00

Related topic

AlternativeDefinitionOfGroup

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