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From: herbzet on 3 Jun 2010 00:24


> Arturo Magidin wrote:

> > This is the well-known condition on groups that says that a semigroup
> > (a nonempty set with a binary associative operation) is a group if and
> > only if for every a and b, the equations
> >
> > a = bx and a = yb

This is nice and neat also.

Can you smell the sawdust burning? I'm thinkin'.

--
hz
From: George Greene on 3 Jun 2010 00:28
On Jun 2, 10:23 pm, herbzet <herb...(a)gmail.com> wrote:
> I think that what's going on is that the variables 'x','y','z'
> (from the end of the alphabet) are being instantiated to
> constants 'a','b','c' (from the beginning of the alphabet).

In order to use the universal-generalization-inference rule at all,
you have to come up with a "new" symbol. It does not even matter
whether you call it a constant or a variable. Usually it is thought
of as an arbitrary constant.

> Quantifying over constants is not quite kosher, I think.

Maybe not, but if you are going to prove a universally quantified
statement,
you will probably start with a constant and replace it by a variable
as you
invoke the rule.

From: Arturo Magidin on 3 Jun 2010 00:30
On Jun 2, 10:59 pm, herbzet <herb...(a)gmail.com> wrote:
> herbzet wrote:
> > Arturo Magidin wrote:
> > > herbzet wrote:
> > > > I think that what we want for a third axiom is
>
> > > > (A3') Axyz [(x.y = x.z) -> (y = z)].
>
> > > This may suffice (I haven't checked), but the original A3 is perfectly
> > > fine too.
>
> (A3') reflects that I'm thinking of a group as a Latin square:
>
>   a Latin square is an n × n table filled with n different symbols
>   in such a way that each symbol occurs exactly once in each row and
>   exactly once in each column (Wikipedia).
>
> except I'm allowing the square to have an infinite number of rows
> and columns.
>
> So I'm conceptualizing a group as a (possibly infinite) Latin square
> (the group operation matrix) that obeys the associative law.
>
> I don't know if that's right, but it seems nice and neat.

By itself, of course it does not suffice (while the Cayley table of
every (finite) group is a (finite) latin square, not every (finite)
latin square is the Cayley table of a (finite) group). My only caveat
was wether together with left-solutions it would suffice. (Note that
if you replaced your (A3') with the similar

(A3'') Axyz [yx=zx -> y=z]

then my semigroup example would satisfy (A1), (A2), and (A3''), and
not be a group. But we can think of this A3'' as being a similarly
flavored "latin square" axiom; in conjunction with A2 it says that the
solution in A2 is unique, so that every symbol will occur in each
column/row [depending how you interpret the first factor] *exactly*
once).

But my example does not satisfy A1, A2, and A3'. So the question is
whether associativity, left cancellation, and solutions on the left
imply a group structure. And I've been wavering, but haven't been able
to come up with a counterexample (and it's getting late). I'll think
about it a bit more.

--
Arturo Magidin
From: herbzet on 3 Jun 2010 03:53


Arturo Magidin wrote:
> herbzet wrote:
> > herbzet wrote:
> > > Arturo Magidin wrote:
> > > > herbzet wrote:
> > > > > I think that what we want for a third axiom is
> >
> > > > > (A3') Axyz [(x.y = x.z) -> (y = z)].
> >
> > > > This may suffice (I haven't checked), but the original A3 is perfectly
> > > > fine too.
> >
> > (A3') reflects that I'm thinking of a group as a Latin square:
> >
> > a Latin square is an n � n table filled with n different symbols
> > in such a way that each symbol occurs exactly once in each row and
> > exactly once in each column (Wikipedia).
> >
> > except I'm allowing the square to have an infinite number of rows
> > and columns.
> >
> > So I'm conceptualizing a group as a (possibly infinite) Latin square
> > (the group operation matrix) that obeys the associative law.
> >
> > I don't know if that's right, but it seems nice and neat.
>
> By itself, of course it does not suffice (while the Cayley table of
> every (finite) group is a (finite) latin square, not every (finite)
> latin square is the Cayley table of a (finite) group).

Right -- the associative law must hold too. With that constraint,
I think my description in English is correct, even for infinite
tables -- but I'm not sure.

> My only caveat
> was wether together with left-solutions it would suffice.

Oh, yeah -- I haven't been thinking about left/right issues.
Sloppiness on my part.

> (Note that
> if you replaced your (A3') with the similar
>
> (A3'') Axyz [yx=zx -> y=z]
>
> then my semigroup example would satisfy (A1), (A2), and (A3''), and
> not be a group. But we can think of this A3'' as being a similarly
> flavored "latin square" axiom; in conjunction with A2 it says that the
> solution in A2 is unique, so that every symbol will occur in each
> column/row [depending how you interpret the first factor] *exactly*
> once).

> But my example does not satisfy A1, A2, and A3'. So the question is
> whether associativity, left cancellation, and solutions on the left
> imply a group structure. And I've been wavering, but haven't been able
> to come up with a counterexample (and it's getting late). I'll think
> about it a bit more.

Me too.

--
hz
From: Frederick Williams on 3 Jun 2010 11:55
Arturo Magidin wrote:
>
> On Jun 2, 7:58 am, Frederick Williams <frederick.willia...(a)tesco.net>
> wrote:
> > Mitchell Hockley wrote:

> >
> > > (Hint: Use Ez(c = z.a))
> >
> > Eh? What if a = 0 and c =/= 0?
>
> I think you are confused; when we use "a=0" in a group, the operation
> is usually taken to be addition, not multiplication.

Yes, sorry. I'm an idiot.

--
I can't go on, I'll go on.
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