Prev: Help with a question - throey of groups using functions
Next: Theory of groups with functions - Help on a question
From: William Elliot on 2 Jun 2010 21:59
On Wed, 2 Jun 2010, Mitchell Hockley wrote:

> "The theory of groups can be presented as having in its vocabulary
> just identity and a single two-place function f(x,y) which we write as
> 'x.y'. The usual laws for identity apply, and in addition these three
> axioms:"
>
> (A1) Axyz(x.(y.z) = (x.y).z)
>
> (A2) AxyEz(x = z.y)
>
> (A3) AxyEz(x = y.z)
>
> Prove:
>
> a = a.c |= c = c.c
>
> (Hint: Use Ez(c = z.a))

Multiply both sides of a = ac, on the left by z.

> Thanks for any help, Mitch.
>
From: herbzet on 2 Jun 2010 22:23


George Greene wrote:
> Arturo Magidin wrote:

> > To the OP: suppose you have a structure satisfying (A1)-(A3), and in
> > addition that Eac(a.c=a)
>
> This is sort of a misuse of the existential quantifier.

Sort of a slight abuse of notation -- it might have been
more correct to just say "and in addition that a.c=a".

But that might have incorrectly suggested to the OP that
A.M. was trying to say that Axy(x.y=x) -- see below.

> You are actually trying to prove that the implication holds
> for ALL a and ALL c.
>
> The original problem is arguably mis-phrased (in the book).
> What you ACTUALLY want to prove is NOT
> > a = a.c |= c = c.c
> BUT RATHER
> Aac[ a=a.c --> c=c.c ]
>
> The fact that the original is stated withOUT quantifiers really
> is problematic unless the book has a convention regarding
> how they ought to be put back in.

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).

Quantifying over constants is not quite kosher, I think.

--
hz
From: Arturo Magidin on 2 Jun 2010 22:40
On Jun 2, 9:21 pm, herbzet <herb...(a)gmail.com> wrote:
> Arturo Magidin wrote:
> > Frederick Williams wrote:
> > > Mitchell Hockley wrote:
>
> > > > Hi,
>
> > > > Could anyone help me with this question (from Bostock's Intermediate
> > > > Logic)
>
> > > > "The theory of groups can be presented as having in its vocabulary
> > > > just identity and a single two-place function f(x,y) which we write as
> > > > 'x.y'. The usual laws for identity apply, and in addition these three
> > > > axioms:"
>
> > > > (A1) Axyz(x.(y.z) = (x.y).z)
>
> Right -- the group operation is associative.
>
> > > > (A2) AxyEz(x = z.y)
>
> Right -- my personal name for this axiom is "the reach axiom" -- because
> you can "reach" any x from any y in one step.
>
> > > > (A3) AxyEz(x = y.z)
>
> Can this be right?  

Yes.

>Taken together with A2, this says that for any x and y,
> there is a z such that x = y.z = z.y.

No, because the "z" that exists by A2 need not be the same "z' that
exists by A3. It says that for any x and for any y, there is a z and a
w such that x = y.z = w.y.

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

both have solutions.

You *do* need both. To see an example where having A2 but not A3 does
not suffice, consider the semigroup with underlying set S with at
least two elements x and y, x=/=y, and operation a.b = a for all a
and b. This operation is trivially associative, and for all a and b
there exists c such that a = c.b, namely c=a. However, this is not a
group because you have x.y = x.x, but x=/=y.


> This would clearly be true in an
> Abelian group, where Axy(x.y = y.x), but is it generally true for all
> groups?  I don't think so.

You are misinterpreting the two existentials. What you have is

(*) AxyEz(x=z.y) /\ AxyEz(x=y.z)

This is *different* from

(**) Axy Ez (x=z.y /\ x=y.z).

(**) is strictly stronger than (*). In (**), the first "Ez" only
applies to the first formula; it is formally equivalent to

(*') AxyEz(x=z.y) /\ ArsEt(r=s.t).

> 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.

--
Arturo Magidin
From: herbzet on 2 Jun 2010 23:16


Arturo Magidin wrote:
> herbzet wrote:
> > Arturo Magidin wrote:
> > > Frederick Williams wrote:
> > > > Mitchell Hockley wrote:
> >
> > > > > Hi,
> >
> > > > > Could anyone help me with this question (from Bostock's Intermediate
> > > > > Logic)
> >
> > > > > "The theory of groups can be presented as having in its vocabulary
> > > > > just identity and a single two-place function f(x,y) which we write as
> > > > > 'x.y'. The usual laws for identity apply, and in addition these three
> > > > > axioms:"
> >
> > > > > (A1) Axyz(x.(y.z) = (x.y).z)
> >
> > Right -- the group operation is associative.
> >
> > > > > (A2) AxyEz(x = z.y)
> >
> > Right -- my personal name for this axiom is "the reach axiom" -- because
> > you can "reach" any x from any y in one step.
> >
> > > > > (A3) AxyEz(x = y.z)
> >
> > Can this be right?
>
> Yes.
>
> >Taken together with A2, this says that for any x and y,
> > there is a z such that x = y.z = z.y.
>
> No, because the "z" that exists by A2 need not be the same "z' that
> exists by A3. It says that for any x and for any y, there is a z and a
> w such that x = y.z = w.y.

Makes sense.

> 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
>
> both have solutions.
>
> You *do* need both. To see an example where having A2 but not A3 does
> not suffice, consider the semigroup with underlying set S with at
> least two elements x and y, x=/=y, and operation a.b = a for all a
> and b. This operation is trivially associative, and for all a and b
> there exists c such that a = c.b, namely c=a. However, this is not a
> group because you have x.y = x.x, but x=/=y.
>
> > This would clearly be true in an
> > Abelian group, where Axy(x.y = y.x), but is it generally true for all
> > groups? I don't think so.
>
> You are misinterpreting the two existentials. What you have is
>
> (*) AxyEz(x=z.y) /\ AxyEz(x=y.z)
>
> This is *different* from
>
> (**) Axy Ez (x=z.y /\ x=y.z).
>
> (**) is strictly stronger than (*). In (**), the first "Ez" only
> applies to the first formula; it is formally equivalent to
>
> (*') AxyEz(x=z.y) /\ ArsEt(r=s.t).
>
> > 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.

Thanks very much.

--
hz
From: herbzet on 2 Jun 2010 23:59


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.

--
hz
First  |  Prev  |  Next  |  Last
Pages: 1 2 3 4 5
Prev: Help with a question - throey of groups using functions
Next: Theory of groups with functions - Help on a question