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compass and straightedge construction

Defines: 
compass, straightedge, ruler, constructible, collapsible compass, collapsible
Synonym: 
straightedge and compass construction, ruler and compass construction, compass and ruler construction
Major Section: 
Reference
Type of Math Object: 
Definition
Groups audience: 

Mathematics Subject Classification

01A20 no label found51M15 no label found

Comments

This is an attempt to reduce the standard axiom set for group.
Let S be a set with an associative law defined everywhere: for any a and b in S, there exists an unique c such that ab = c.
This associative law satisfies the following axioms:

1 - There exists at least one right-neutral element e (may be more f, g...) such that ae = a for all a in S (af=a, ag=a...)

2 - Every a in S has at least one right inverse a' (maybe more) with respect to one of the right elements e, such that aa' = e.

Axiom 2 differs from the standard definition: for a and b in S, there are right inverse a' and b', not necessarily unique, such that aa' = e and bb' = f, but e and f could be different right-neutral elements.

Is such a set still a group?

Here's what I think:

1. Under the proposed axiom set, there exists at most one right inverse, but there could exist none, for you only say the right-hand side of $ab=c$ exists. If I ask you to construct $x^{-1}$ for some $x$ in the group, what $a$ and $b$ do you take?

What if the set is the integers with multiplication?
For any integer $a$ and $b$ there certainly is another unique integer $c$ such that $ab=c$. e.g. $2 \times 3 = 6$, but $2 \times 3 \neq 4$.

2. Along the same lines, what is the procedure for constructing the identity $e$? Well, you might say $x x^{-1} = e$, but this is of course circular with (1).

3. The set could be empty.

// Steve

Sorry, I spoke too soon ... :(
Please ignore the nonsense that I just posted.

I think I had misread your post as saying that associative law is proving can be used to prove the two axioms that you list.

// Steve

> Let S be a set with an associative law defined everywhere:

Do you mean a binary operation on S satisfying the associative law?

> This associative law satisfies the following axioms:
>
> 1 - There exists at least one right-neutral element e (may
> be more f, g...) such that ae = a for all a in S (af=a,
> ag=a...)
>
> 2 - Every a in S has at least one right inverse a' (maybe
> more) with respect to one of the right elements e, such that
> aa' = e.
>

Do you mean there is a "right-neutral element" e such that, for each a in S, there is a' in S such that aa'=e?

If "minimizing" the group axiom set means "minimizing" the number of identities required to define a "group", then it has been shown that a group is equivalent to an algebraic system whose underlying set is the underlying set of the group, with one binary operation called "division" and one identity.

See ftp://info.mcs.anl.gov/pub/tech_reports/reports/P901.pdf and http://www.cs.unm.edu/~mccune/projects/gtsax/

for more details.

No, these two articles deal with a method of writing a given set of axioms in a more compact form. What I am presenting is a different set of axioms. Maybe the term "miminizing" is not appropriate, I meant in fact, a weaker set of axioms.

Standard set of axioms for the law of a group G:
1 - The law is associative.
2 - There exists a neutral element e sucha that ae = ea = a for all a in G.
3 - All a in G have an inverse a' such that aa' = a'a = e.

"Weak" set of axioms:
1 - The law is associative.
2 - There exists a right-neutral element e such that ae = a for all a in G.
3 - All a in G have a right inverse a' such that aa' = e.

It can be proven that the weak set is equivalent to the "stong" set by considering the product aa'a", where a" is the right inverse of a'. Then, we get the following results:
I - a right inverse is also a left inverse.
II - the right-neutral element is also left-neutral
III - the right inverse is unique, as is the neutral element.

What I suggest is to try to make axiom 3 even weaker: the inverses of a', b' of a and b are such that aa' = e and bb' = f, and the righ-elements e and f could be different. If we try to apply the former method, we get a problem:
a' being the right inverse of a, aa' = e. If a" is the right inverse of a', a'a" = f and f could be different from e. From the product aa'a" we get the following results:
I - a'a = f. The right inverse is also left inverse but with respect to another right-neutral element.
II - ea = a. The right-neutral element a is also left-neutral for a only. if bb' = f for example, then f will be left-neutral for b and not for a.

The question is: is there a way to complete the proof and get the standard set of axioms?

dh2718 writes:

> The question is: is there a way to complete
> the proof and get the standard set of axioms?

No. On any non-empty set, define multiplication by xy = x. This satisfies your axioms, but if the set has more than one element it's not a group.

Good counter-example, end of the story.

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