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Mar 4, 1986, 9:37:40 AM3/4/86

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If we modify the definition of a group slightly, do new groups arise?

I have never had a group theory course, so if this is all standard established

knowledge, forgive me. I thought the question was interesting.

Review: A group is a set of elements S and an operator '*',

possessing the following properties:

The binary operator '*' is a well defined function, closed in S.

The operator '*' is associative.

There is a unique identity element E, such that X*E = E*X = X for every X.

Every X has a unique inverse Y, such that X*Y = Y*X = E.

Consider the identity and inverse properties of a group, as outlined above.

Suppose the order of the operands is significant.

Example: every X has a unique inverse Y, such that X*Y = E.

No constraints placed upon Y*X.

There are four possible "weakened" definitions:

identity inverse

1. X*E = X X*Y = E

2. E*X = X X*Y = E

3. E*X = X Y*X = E

4. X*E = X Y*X = E

If I have not made any serious blunders,

each weaker definition implies the other three,

resurrecting the original concept.

By symmetry, definitions 1 and 3 are equivalent, producing reflected theorems.

Similarly, 2 and 4 are equivalent. Therefore, we consider only 1 and 2.

Suppose we adopt definition 1.

Every X has an inverse Y, and this Y must have an inverse Z.

Multiply the equation X*Y = E by Y on the left and Z on the right.

After invoking associativity and identity properties,

we have Y*X*(Y*Z) = Y*Z.

Since Z is the inverse of Y, this reduces to Y*X = E.

Thus, if Y is the inverse of X, then X is the inverse of Y.

To prove E*X = X, begin with Y*E = Y, where Y is the inverse of X.

Multiply by X on the right and on the left, and reduce the equation to E*X = X.

Thus, definition 1 implies all the original axioms.

Starting with definition 2 is more challenging.

I will leave this as an exercise for the interested.

--

When the sky becomes uranious, we will all go simultaneous.

Karl Dahlke ihnp4!ihnet!eklhad

Mar 7, 1986, 4:49:33 PM3/7/86

to

In article <3...@ihnet.UUCP>, ekl...@ihnet.UUCP (K. A. Dahlke) writes:

> If we modify the definition of a group slightly, do new groups arise?

> If we modify the definition of a group slightly, do new groups arise?

You might be interested in loops, where the multiplication has

different left and right inverses. Unfortunately, elementary

accounts are rare. There is a volume in the Springer Ergebnisse

series on loops, written by Bruck. They're useful for representing

Latin Squares. There's a more recent reference which currently

escapes me.

Peter Ladkin

Mar 7, 1986, 8:06:00 PM3/7/86

to

Re: Groups, Alternative Definitions?

Definitions 2 and 4 are not equivalent to the definition

of a group. Here's a well-known example (see, for instance,

Fraleigh's abstract algebra book):

Let S be the set of nonzero real numbers and for

a and b in S define a*b to be |a| times b ( |a| = absolute

value of a ). This operation is well-defined and associative.

Furthermore, it satisfies Definition 2 (take E = 1 and

Y = |X|^(-1) ). However, we can readily check that (S,*) is

not a group. For the only values of E for which E*X = X for

all X in S are E = 1 or E = -1. Neither value satisfies

X*E = X for all X in S (for E = 1, take any X < 0 and for

E = -1, take any X > 0). Thus, (S,*) has no identity and is

not a group.

Michael Ward UUCP . . .!tektronix!uoregon!mward

University of CSNET mward@uoregon

Oregon

Mar 8, 1986, 1:26:48 PM3/8/86

to rnews@mcvax

In article <3...@ihnet.UUCP> ekl...@ihnet.UUCP writes:

> If we modify the definition of a group slightly, do new groups arise?

> [...]> If we modify the definition of a group slightly, do new groups arise?

> There are four possible "weakened" definitions:

> identity inverse

> 1. X*E = X X*Y = E

> 2. E*X = X X*Y = E

> If I have not made any serious blunders, each weaker definition implies the

> other three, resurrecting the original concept.

There is an interesting difference between 1, which has the unit and

inverse on the same side of the * operation, and 2, where unit and inverse

are on different sides. For case 1, it is not necessary to assume

uniqueness of unit and inverse; the existence of at least one unit for the

whole "group" and at least one inverse for each element is sufficient.

Uniqueness can then be proved.

For 2, we only get a group if we assume uniqueness of the unit E. If we

just assume the existence of *some* unit E, a counterexample is found by

defining X*Y = Y. This operation is associative: (X*Y)*Z = Z = Y*Z =

X*(Y*Z). Each element is a left unit. Take an arbitrary one and call it

E. Then * satisfies both axioms in 2: E*X = X, and X*Y = E for Y = E.

(All this is well known; at least it was part of the first lecture in an

introductory algebra class I once took.)

--

Lambert Meertens

...!{seismo,okstate,garfield,decvax,philabs}!lam...@mcvax.UUCP

CWI (Centre for Mathematics and Computer Science), Amsterdam

Mar 8, 1986, 6:01:28 PM3/8/86

to

Karl Dahlke observes that the group axioms need only specify

right- (or left-) inverses instead of two-sided inverses, since

the latter may be deduced from the former. There are other

possible formulations of the group axioms. For example,

uniqueness of inverse can be dropped since it follows as a

consequence of the weaker formulation anyway. Textbook

presentations of group theory (e.g. Herstein) usually have

several exercises in which one is to show that an apparently

weaker set of axioms for a (set,operator) is sufficient to

make the (set,operator) constitute a group.

right- (or left-) inverses instead of two-sided inverses, since

the latter may be deduced from the former. There are other

possible formulations of the group axioms. For example,

uniqueness of inverse can be dropped since it follows as a

consequence of the weaker formulation anyway. Textbook

presentations of group theory (e.g. Herstein) usually have

several exercises in which one is to show that an apparently

weaker set of axioms for a (set,operator) is sufficient to

make the (set,operator) constitute a group.

Group theory news item: The latest BAMS contains a tutorial

article by Gorenstein on the group classification theorem.

Mar 10, 1986, 1:43:00 PM3/10/86

to

In my earlier response to Karl Dahlke's alternative

formulations of the group axioms I carelessly overlooked

the word "unique" in his definitions. The example I

mentioned shows that without the assumption of the uniqueness

of the left identity, the existence of a left identity and right

inverses does not always yield a group. In the presence of

the uniqueness assumption on the left identity, a group is

obtained.

Michael Ward UUCP . . .!tektronix!uoregon!mward

CSNET mward@uoregon

Mar 12, 1986, 2:24:22 PM3/12/86

to

> You might be interested in loops, where the multiplication has

> different left and right inverses. Unfortunately, elementary

> accounts are rare. There is a volume in the Springer Ergebnisse

> series on loops, written by Bruck. They're useful for representing

> Latin Squares. There's a more recent reference which currently

> escapes me.

> different left and right inverses. Unfortunately, elementary

> accounts are rare. There is a volume in the Springer Ergebnisse

> series on loops, written by Bruck. They're useful for representing

> Latin Squares. There's a more recent reference which currently

> escapes me.

The last book I read on Latin Squares called them "quasigroups" or

something like that. "Loops" sounds much better--if they're indeed

the same.

"An antiset is a mathematical object that fails to

contain each of its members." --E. P. B. Umbugio

--

Col. G. L. Sicherman

UU: ...{rocksvax|decvax}!sunybcs!colonel

CS: colonel@buffalo-cs

BI: csdsicher@sunyabva

Mar 16, 1986, 3:19:57 PM3/16/86

to

In article <8...@ellie.UUCP> ellie!colonel (Col. G. L. Sicherman) writes:

>The last book I read on Latin Squares called them "quasigroups" or

>something like that. "Loops" sounds much better--if they're indeed

>the same.

>The last book I read on Latin Squares called them "quasigroups" or

>something like that. "Loops" sounds much better--if they're indeed

>the same.

If I remember correctly, a loop is a quasigroup with an identity element.

The quasigroup axioms include cancellation (if xz=yz or zx=zy then x=y)

but not associativity; an associative quasigroup (or loop) is a group.

Another way to say this is that an object which is both a quasigroup and

a semigroup is a group.

What's the origin of the name "Loop"? Is it a contraction of "Latin-

square group", or what? (I see nothing that suggests roundness in either

loops or rings.)

Mar 17, 1986, 4:07:58 PM3/17/86

to

In article <8...@ellie.UUCP>, col...@ellie.UUCP (Col. G. L. Sicherman) writes:

>

> The last book I read on Latin Squares called them "quasigroups" or

> something like that. "Loops" sounds much better--if they're indeed

> the same.

>

> The last book I read on Latin Squares called them "quasigroups" or

> something like that. "Loops" sounds much better--if they're indeed

> the same.

Whoops! I guess I don't have my definitions right. They're very

close, but not the same. I think one has an identity and one not.

Peter Ladkin

Mar 19, 1986, 11:33:38 PM3/19/86

to

[Balanced meal for line eater goes here.]

As you may know, the Mandelbrot set is the set of complex numbers c

such that the sequence

2 3

0, f(0), f (0), f (0), ...

k

is not bounded, where f(z) = z^2+c, and f (z) = f(f(f(...f(z)...)))

k

(k applications of f). It is claimed that if |f (0)| > 2 for some

k, then the seqence blows up (in absolute value) to infinity. Can

someone show a proof for this?

Graeme Clark -- Dept. of Computer Science, Univ. of Toronto, Canada M5S 1A4

{allegra,cornell,decvax,ihnp4,linus,utzoo}!utcsri!gclark

Mar 28, 1986, 12:28:58 PM3/28/86

to

Let x(0) = 0, x(k + 1) = x(k) ^ 2 - a. Denote the absolute value

function abs. Then

Theorem: If abs(x(n)) > 2 for some n, then lim abs(x(k)) = inf.

Lemma: If

x(k) > (1 + sqrt(1 + 4*abs(a))) / 2,

then lim abs(x(k)) = inf.

Proof: The above inequality implies

x(k) > e + (1 + sqrt(1 + 4*abs(a))) / 2

for some e > 0. By the triangle inequality,

abs(x(k + 1)) = abs(x(k) ^ 2 - a) >= abs(x(k)) ^ 2 - abs(a) >

abs(x(k)) + e*sqrt(1 + 4*abs(a)).

By induction,

abs(x(k + N)) > abs(x(k)) + N*e.

The lemma follows.

Proof: If abs(x(n)) = b > 2 for some n, then

max(b, abs(x(1))) >= (2*b + abs(x(1))) / 3 >

(4 + abs(a)) / 3 >= (1 + sqrt(1 + 4*abs(a))) / 2.

The theorem follows immediately from the lemma.

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