Transitive groups of degree 12 and low order containing a 12-cycle

Here are generators for some of these; since they contain a 12-cycle
the chords of these systems are transposible, and they might even make
sense in musical terms.

Groups containing 12-cycle

C(4) x C(3) = C(12)
Order 12

a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]

S(3) x C(4)
Order 24

a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
b := [[1, 5], [2, 10], [4, 8], [7, 11]]
e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]

1/2[3:2]cD(4) = D(12)
Order 24

a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
w := [[1, 11], [2, 10], [3, 9], [4, 8], [5, 7]]

D(4) x C(3)
Order 24

a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
k := [[1, 7], [3, 9], [5, 11]]

[3^2]4
Order 36

e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
kk := [[12, 4, 8], [2, 6, 10]]

[3^2]4'
Order 36

e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
r := [[12, 4, 8], [2, 6, 10]]

D(4) x S(3)
Order 48

a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
b := [[1, 5], [2, 10], [4, 8], [7, 11]]
e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
k := [[1, 7], [3, 9], [5, 11]]

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

Here are the polynomials for these, giving the number chords of each size

> C(4) x C(3) = C(12)
> Order 12

x^12+x^11+6*x^10+19*x^9+43*x^8+66*x^7+80*x^6+66*x^5+43*x^4+19*x^3+6*x^2+x+1

> S(3) x C(4)
> Order 24

x^12+x^11+5*x^10+12*x^9+28*x^8+38*x^7+48*x^6+38*x^5+28*x^4+12*x^3+5*x^2+x+1

> 1/2[3:2]cD(4) = D(12)
> Order 24

x^12+x^11+6*x^10+12*x^9+29*x^8+38*x^7+50*x^6+38*x^5+29*x^4+12*x^3+6*x^2+x+1

> D(4) x C(3)
> Order 24

x^12+x^11+5*x^10+13*x^9+28*x^8+40*x^7+50*x^6+40*x^5+28*x^4+13*x^3+5*x^2+x+1

> [3^2]4
> Order 36

x^12+x^11+5*x^10+9*x^9+22*x^8+26*x^7+36*x^6+26*x^5+22*x^4+9*x^3+5*x^2+x+1

The generators of this were given incorrectly, it should have been:

e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
kk := [[12, 4, 8], [2, 10, 10]]

> [3^2]4'
> Order 36

x^12+x^11+4*x^10+9*x^9+18*x^8+26*x^7+32*x^6+26*x^5+18*x^4+9*x^3+4*x^2+x+1

As distinct from:

> e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> r := [[12, 4, 8], [2, 6, 10]]

> D(4) x S(3)
> Order 48

x^12+x^11+5*x^10+9*x^9+21*x^8+25*x^7+34*x^6+25*x^5+21*x^4+9*x^3+5*x^2+x+1

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
>
> Here are the polynomials for these, giving the number chords of
each size
>
> > C(4) x C(3) = C(12)
> > Order 12
>
>
x^12+x^11+6*x^10+19*x^9+43*x^8+66*x^7+80*x^6+66*x^5+43*x^4+19*x^3+6*x^
2+x+1
>
> > S(3) x C(4)
> > Order 24
>
>
x^12+x^11+5*x^10+12*x^9+28*x^8+38*x^7+48*x^6+38*x^5+28*x^4+12*x^3+5*x^
2+x+1
>
> > 1/2[3:2]cD(4) = D(12)
> > Order 24
>
>
x^12+x^11+6*x^10+12*x^9+29*x^8+38*x^7+50*x^6+38*x^5+29*x^4+12*x^3+6*x^
2+x+1
>
> > D(4) x C(3)
> > Order 24
>
>
x^12+x^11+5*x^10+13*x^9+28*x^8+40*x^7+50*x^6+40*x^5+28*x^4+13*x^3+5*x^
2+x+1
>
> > [3^2]4
> > Order 36
>
>
x^12+x^11+5*x^10+9*x^9+22*x^8+26*x^7+36*x^6+26*x^5+22*x^4+9*x^3+5*x^2+
x+1
>
> The generators of this were given incorrectly, it should have been:
>
> e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> kk := [[12, 4, 8], [2, 10, 10]]
>
> > [3^2]4'
> > Order 36
>
>
x^12+x^11+4*x^10+9*x^9+18*x^8+26*x^7+32*x^6+26*x^5+18*x^4+9*x^3+4*x^2+
x+1
>
> As distinct from:
>
> > e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> > r := [[12, 4, 8], [2, 6, 10]]
>
>
> > D(4) x S(3)
> > Order 48
>
>
x^12+x^11+5*x^10+9*x^9+21*x^8+25*x^7+34*x^6+25*x^5+21*x^4+9*x^3+5*x^2+
x+1

I recognize the first and third polynomials as being counts for 12-et
Tn-types and TnI-types respectively (for 12 down to 1). Is this just
a coincidence? Do these polynomials (also) have another function?

Thanks

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> I recognize the first and third polynomials as being counts for 12-
et
> Tn-types and TnI-types respectively (for 12 down to 1). Is this
just
> a coincidence? Do these polynomials (also) have another function?

Not at all. What you are calling Tn types and TnI types should
probably be called C(12) types and D(12) types or something of that
sort, since this explicitly names the permutation group in question.
What is Tn short for?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > I recognize the first and third polynomials as being counts for
12-
> et
> > Tn-types and TnI-types respectively (for 12 down to 1). Is this
> just
> > a coincidence? Do these polynomials (also) have another function?
>
> Not at all. What you are calling Tn types and TnI types should
> probably be called C(12) types and D(12) types or something of that
> sort, since this explicitly names the permutation group in question.
> What is Tn short for?

This is John Rahn's notation. It's actually T(sub)n and T(sub)n I
It stands for Transpose, n is the amount in halfsteps I is Inversion
(for mirror images). What application do the other polynomials have
to music theory?

Thanks

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

>What application do the other polynomials have
> to music theory?

They enumerate distinct 12-equal chord forms under other permutation
groups. For instance, we might be interested in enumnerating
disctinct chords under D(4) x S(3); this group is generated by the
cyclic permutation giving transpositions, inversion, and the
operation of converting the circle of semitones to a circle of
fifths, by sending the ith note to 7i mod 12.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> >What application do the other polynomials have
> > to music theory?
>
> They enumerate distinct 12-equal chord forms under other
permutation
> groups. For instance, we might be interested in enumnerating
> disctinct chords under D(4) x S(3); this group is generated by the
> cyclic permutation giving transpositions, inversion, and the
> operation of converting the circle of semitones to a circle of
> fifths, by sending the ith note to 7i mod 12.

I think you've touched on something that was just being discussed on
this list! Though I wasn't following very closely . . .

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> >What application do the other polynomials have
> > to music theory?
>
> They enumerate distinct 12-equal chord forms under other
permutation
> groups. For instance, we might be interested in enumnerating
> disctinct chords under D(4) x S(3); this group is generated by the
> cyclic permutation giving transpositions, inversion, and the
> operation of converting the circle of semitones to a circle of
> fifths, by sending the ith note to 7i mod 12.

Cool. This was what Jon Wild was talking about. I would like to see
how one works this into Polya's algorithm. I was introduced to this
One-Seven symmetry by John Rahn, its neat to see the group-theoretical
basis for this (D(4)XS(3)) Thanks

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> Cool. This was what Jon Wild was talking about. I would like to see
> how one works this into Polya's algorithm. I was introduced to this
> One-Seven symmetry by John Rahn, its neat to see the group-
theoretical
> basis for this (D(4)XS(3)) Thanks

I gave the polynomial for it in another posting. If we have a
permutation group, then for each element we form a product of
(x^d+1) for each d-cycle; the degree of the resulting polynomial is
necesarily the degee of the permutation group. Dividing by the order
of the group gives the polynomial.

There is something similar I've long worked with, called the Molien
series, by the way, where instead of (x^d+1) one takes 1/(1-x^d);
expanding the resulting rational function in a power series gives the
number of invariants of each degree m as the coefficient of x^m.
That, I suppose, might have its uses also.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> > wrote:
> >
> > Here are the polynomials for these, giving the number chords of
> each size
> >
> > > C(4) x C(3) = C(12)
> > > Order 12

* What do C and S stand for? (I've tried Eric Weisstein's site...)
> >
> >
>
x^12+x^11+6*x^10+19*x^9+43*x^8+66*x^7+80*x^6+66*x^5+43*x^4+19*x^3+6*x^
> 2+x+1
> >
> > > S(3) x C(4)
> > > Order 24
> >
> >
>
x^12+x^11+5*x^10+12*x^9+28*x^8+38*x^7+48*x^6+38*x^5+28*x^4+12*x^3+5*x^
> 2+x+1
> >
> > > 1/2[3:2]cD(4) = D(12)

* What does the above stand for?

> > > Order 24
> >
> >
>
x^12+x^11+6*x^10+12*x^9+29*x^8+38*x^7+50*x^6+38*x^5+29*x^4+12*x^3+6*x^
> 2+x+1
> >
Thanks

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> > > > C(4) x C(3) = C(12)
> > > > Order 12
>
> * What do C and S stand for? (I've tried Eric Weisstein's site...)

C(n) is the cyclic group of degree and order n, and S(n) is the
symmetric group of degree n.

> > > > 1/2[3:2]cD(4) = D(12)
>
> * What does the above stand for?

D(12) is the dihedral group of degree 12 (and order 24)--the group of
the dodecagon.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > > > > C(4) x C(3) = C(12)
> > > > > Order 12
> >
> > * What do C and S stand for? (I've tried Eric Weisstein's site...)
>
> C(n) is the cyclic group of degree and order n, and S(n) is the
> symmetric group of degree n.
>
> > > > > 1/2[3:2]cD(4) = D(12)
> >
> > * What does the above stand for?
>
> D(12) is the dihedral group of degree 12 (and order 24)--the group
of
> the dodecagon.

Thanks - I'm working to figure out why D(4)X S(3) is the same as
counting sets after reducing for the 1-5 symmetry (1,5,7,11 act)

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul.hjelmstad@u...> wrote:
> >
> > > > > > C(4) x C(3) = C(12)
> > > > > > Order 12
> > >
> > > * What do C and S stand for? (I've tried Eric Weisstein's
site...)
> >
> > C(n) is the cyclic group of degree and order n, and S(n) is the
> > symmetric group of degree n.
> >
> > > > > > 1/2[3:2]cD(4) = D(12)
> > >
> > > * What does the above stand for?
> >
> > D(12) is the dihedral group of degree 12 (and order 24)--the
group
> of
> > the dodecagon.
>
> Thanks - I'm working to figure out why D(4)X S(3) is the same as
> counting sets after reducing for the 1-5 symmetry (1,5,7,11 act)

I've been thinking about this. If you map the 12 pitches of 12-et
in a 4 X 3 grid thus:

0 3 6 9
4 7 10 1
8 11 2 5

Selecting any subset of 12-et (coloring the squares for example)
If you flip the whole thing along a vertical axis in the middle,
you will get the P1<->P7 transform. Flipping along a horizontal
axis in the middle will give the mirror image and P1<->P7 transform.
(rotating the colors 180 degrees gives the mirror image only)
So it makes sense that this is D(4) X S(3). Gene, is this
illustration any good? I'm still kind of figuring out the distinction
between Dihedral and Cyclic groups.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"

> I've been thinking about this. If you map the 12 pitches of 12-et
> in a 4 X 3 grid thus:
>
> 0 3 6 9
> 4 7 10 1
> 8 11 2 5
>
> Selecting any subset of 12-et (coloring the squares for example)
> If you flip the whole thing along a vertical axis in the middle,
> you will get the P1<->P7 transform. Flipping along a horizontal
> axis in the middle will give the mirror image and P1<->P7 transform.
> (rotating the colors 180 degrees gives the mirror image only)
> So it makes sense that this is D(4) X S(3). Gene, is this
> illustration any good? I'm still kind of figuring out the
distinction
> between Dihedral and Cyclic groups.

I like it, but for the S(3) part you need to be willing to permute
the three rows in any order.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
>
> > I've been thinking about this. If you map the 12 pitches of 12-et
> > in a 4 X 3 grid thus:
> >
> > 0 3 6 9
> > 4 7 10 1
> > 8 11 2 5
> >
> > Selecting any subset of 12-et (coloring the squares for example)
> > If you flip the whole thing along a vertical axis in the middle,
> > you will get the P1<->P7 transform. Flipping along a horizontal
> > axis in the middle will give the mirror image and P1<->P7
transform.
> > (rotating the colors 180 degrees gives the mirror image only)
> > So it makes sense that this is D(4) X S(3). Gene, is this
> > illustration any good? I'm still kind of figuring out the
> distinction
> > between Dihedral and Cyclic groups.
>
> I like it, but for the S(3) part you need to be willing to permute
> the three rows in any order.

Actually, you can. Do not forget the cyclical component!
With rotations and inversions together you can get any row order,
because D3 = S3. (This does not hold for D4 and hence does not work
for the columns.)

Hans Straub

--- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> >
> > > I've been thinking about this. If you map the 12 pitches of 12-
et
> > > in a 4 X 3 grid thus:
> > >
> > > 0 3 6 9
> > > 4 7 10 1
> > > 8 11 2 5
> > >
> > > Selecting any subset of 12-et (coloring the squares for example)
> > > If you flip the whole thing along a vertical axis in the middle,
> > > you will get the P1<->P7 transform. Flipping along a horizontal
> > > axis in the middle will give the mirror image and P1<->P7
> transform.
> > > (rotating the colors 180 degrees gives the mirror image only)
> > > So it makes sense that this is D(4) X S(3). Gene, is this
> > > illustration any good? I'm still kind of figuring out the
> > distinction
> > > between Dihedral and Cyclic groups.
> >
> > I like it, but for the S(3) part you need to be willing to
permute
> > the three rows in any order.
>
> Actually, you can. Do not forget the cyclical component!
> With rotations and inversions together you can get any row order,
> because D3 = S3. (This does not hold for D4 and hence does not work
> for the columns.)
>
> Hans Straub

Thanks. Now that I understand Dihedral Groups, let's take the columns
and put them in any order (is there a type of symmetry group for
this?) Anyway, with hexachords, start with D(4)X S(3). This gives
a count of 34 hexachords. Reducing for any column order (Let's call
it F(4). ) This brings the count down to 17 hexachord-classes. It's a
little confusing because I have not reduced for the Z_relation. And
3 set types are Z-related to their P1<->P7 counterpart. Reducing for
the Z-relation brings the count down to 12.

For Pentachords D(4) X S(3) gives 25 chords. Reducing for F(4)X S(3)
gives 13 chords. 3 are Z-related, but since they are Z-related to
their P1<->P7 counterpart there is no further reduction.

For Tetrachords 21 chords reduces to 10. The 1 Z-related tetrachord
is absorbed

Triads, 9 chords reduce down to 6.

Paul

Corrected errors, see at the end. - Paul Hj

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> <gwsmith@s...>
> > wrote:
> > > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > >
> > > > I've been thinking about this. If you map the 12 pitches of
12-
> et
> > > > in a 4 X 3 grid thus:
> > > >
> > > > 0 3 6 9
> > > > 4 7 10 1
> > > > 8 11 2 5
> > > >
> > > > Selecting any subset of 12-et (coloring the squares for
example)
> > > > If you flip the whole thing along a vertical axis in the
middle,
> > > > you will get the P1<->P7 transform. Flipping along a
horizontal
> > > > axis in the middle will give the mirror image and P1<->P7
> > transform.
> > > > (rotating the colors 180 degrees gives the mirror image only)
> > > > So it makes sense that this is D(4) X S(3). Gene, is this
> > > > illustration any good? I'm still kind of figuring out the
> > > distinction
> > > > between Dihedral and Cyclic groups.
> > >
> > > I like it, but for the S(3) part you need to be willing to
> permute
> > > the three rows in any order.
> >
> > Actually, you can. Do not forget the cyclical component!
> > With rotations and inversions together you can get any row order,
> > because D3 = S3. (This does not hold for D4 and hence does not
work
> > for the columns.)
> >
> > Hans Straub
>
> Thanks. Now that I understand Dihedral Groups, let's take the
columns
> and put them in any order (is there a type of symmetry group for
> this?) Anyway, with hexachords, start with D(4)X S(3). This gives
> a count of 34 hexachords. Reducing for any column order (Let's call
> it F(4). ) This brings the count down to 17 hexachord-classes. It's
a
> little confusing because I have not reduced for the Z_relation. And
> 3 set types are Z-related to their P1<->P7 counterpart. Reducing for
> the Z-relation brings the count down to 12.
>
> For Pentachords D(4) X S(3) gives 25 chords. Reducing for F(4)X S(3)
> gives 13 chords. 3 are Z-related, but since they are Z-related to
> their P1<->P7 counterpart there is no further reduction.
>
> For Tetrachords 21 chords reduces to 10. The 1 Z-related tetrachord
> is absorbed
>
> Triads, 9 chords reduce down to 6.

Oops. A couple errors. There is one Z-related Pentachord pair that
isn't a P1<->P7 pair, so Pentachords reduce down to 12 chord classes
after Z-relation-reduction.

Also, Tetrachords reduce down to 11 classes (not 10). Triads are
fine. Diads reduce from 5 chords in S(3) X D(4) down to 3 classes,
when columns are can be interchanged. Monads: there is 1.

So, you get (1,3,6,11,12,12,12,11,6,3,1,1) not counting the empty
set, if you want to reduce for the Z-relation. Otherwise, keeping it
clean from Z-relation-reduction* you obtain
(1,3,6,11,13,17,13,11,6,3,1,1) not counting the empty set.

(I've plotted all these 4 X 3 grids in Visio.)

Paul

* except for Z-relations that are absorbed, of course

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
>
> Thanks. Now that I understand Dihedral Groups, let's take the columns
> and put them in any order (is there a type of symmetry group for
> this?) Anyway, with hexachords, start with D(4)X S(3). This gives
> a count of 34 hexachords. Reducing for any column order (Let's call
> it F(4). )

That is S4, the full symmetric permutation group. I think one strong
point of the diagram is that the actions on the prime components are
directly visible.
But for what purpose do you want to use this special group?

Hans Straub

--- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
> >
> > Thanks. Now that I understand Dihedral Groups, let's take the
columns
> > and put them in any order (is there a type of symmetry group for
> > this?) Anyway, with hexachords, start with D(4)X S(3). This gives
> > a count of 34 hexachords. Reducing for any column order (Let's
call
> > it F(4). )
>
> That is S4, the full symmetric permutation group. I think one strong
> point of the diagram is that the actions on the prime components are
> directly visible.
> But for what purpose do you want to use this special group?
>
> Hans Straub

It's just another way to arrange hexachords (pentachords, etc). There
are so many ways to slice this stuff! Another way to illustrate
hexachords is to use I-Ching symbols: - -, ---, -o-, -e- as follows

0 6
1 7
2 8
3 9
4 10
5 11

Call the complementary hexachords A and A'=B. Then - - is A, B
--- is A, A, -o- is B,A and -e- is B,B. Holding the tritones constant
and varying the others gives some interesting patterns.

Paul

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul.hjelmstad@u...> wrote:
> > >
> > > Thanks. Now that I understand Dihedral Groups, let's take the
> columns
> > > and put them in any order (is there a type of symmetry group
for
> > > this?) Anyway, with hexachords, start with D(4)X S(3). This
gives
> > > a count of 34 hexachords. Reducing for any column order (Let's
> call
> > > it F(4). )
> >
> > That is S4, the full symmetric permutation group. I think one
strong
> > point of the diagram is that the actions on the prime components
are
> > directly visible.
> > But for what purpose do you want to use this special group?
> >
> > Hans Straub
>
> It's just another way to arrange hexachords (pentachords, etc).
There
> are so many ways to slice this stuff! Another way to illustrate
> hexachords is to use I-Ching symbols: - -, ---, -o-, -e- as follows
>
> 0 6
> 1 7
> 2 8
> 3 9
> 4 10
> 5 11
>
> Call the complementary hexachords A and A'=B. a is a member of A
> and b is a member of B. Then - - is (a, b), --- is (a, a), -o- is
>(b,a) and -e- is (b,b). Holding the tritones constant and varying
> the others gives some interesting patterns.
>
> Paul

Gene,

If it's not too much trouble, could you run S(4) X C(4) and also
S(4) X S(3)? And give the generators and polynomials, if its not too
much trouble

Thanx!

Paul Hj

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> Here are generators for some of these; since they contain a 12-cycle
> the chords of these systems are transposible, and they might even
make
> sense in musical terms.
>
> Groups containing 12-cycle
>
> C(4) x C(3) = C(12)
> Order 12
>
> a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
>
>
> S(3) x C(4)
> Order 24
>
> a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> b := [[1, 5], [2, 10], [4, 8], [7, 11]]
> e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
>
> 1/2[3:2]cD(4) = D(12)
> Order 24
>
> a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> w := [[1, 11], [2, 10], [3, 9], [4, 8], [5, 7]]
>
> D(4) x C(3)
> Order 24
>
> a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> k := [[1, 7], [3, 9], [5, 11]]
>
> [3^2]4
> Order 36
>
> e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> kk := [[12, 4, 8], [2, 6, 10]]
>
> [3^2]4'
> Order 36
>
> e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> r := [[12, 4, 8], [2, 6, 10]]
>
>
> D(4) x S(3)
> Order 48
>
> a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> b := [[1, 5], [2, 10], [4, 8], [7, 11]]
> e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> k := [[1, 7], [3, 9], [5, 11]]

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> Gene,
>
> If it's not too much trouble, could you run S(4) X C(3) and also
> S(4) X S(3)? And give the generators and polynomials, if its not
too
> much trouble
>
> Thanx!
>
> Paul Hj
>
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> > Here are generators for some of these; since they contain a 12-
cycle
> > the chords of these systems are transposible, and they might even
> make
> > sense in musical terms.
> >
> > Groups containing 12-cycle
> >
> > C(4) x C(3) = C(12)
> > Order 12
> >
> > a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> > e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> >
> >
> > S(3) x C(4)
> > Order 24
> >
> > a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> > b := [[1, 5], [2, 10], [4, 8], [7, 11]]
> > e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> >
> > 1/2[3:2]cD(4) = D(12)
> > Order 24
> >
> > a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> > e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> > w := [[1, 11], [2, 10], [3, 9], [4, 8], [5, 7]]
> >
> > D(4) x C(3)
> > Order 24
> >
> > a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> > e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> > k := [[1, 7], [3, 9], [5, 11]]
> >
> > [3^2]4
> > Order 36
> >
> > e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> > kk := [[12, 4, 8], [2, 6, 10]]
> >
> > [3^2]4'
> > Order 36
> >
> > e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> > r := [[12, 4, 8], [2, 6, 10]]
> >
> >
> > D(4) x S(3)
> > Order 48
> >
> > a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> > b := [[1, 5], [2, 10], [4, 8], [7, 11]]
> > e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> > k := [[1, 7], [3, 9], [5, 11]]

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> Gene,
>
> If it's not too much trouble, could you run S(4) X C(4) and also
> S(4) X S(3)? And give the generators and polynomials, if its not too
> much trouble

S(4)XC(3) order 72

Generators

a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
B := [[1, 10], [2, 5], [6, 9]]

Polynomial

x^12+x^11+3*x^10+8*x^9+14*x^8+19*x^7+24*x^6+19*x^5+14*x^4+8*x^3+3*x^2+x+1

S(4)XS(3) order 144

Generators

a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
b := [[1, 5], [2, 10], [4, 8], [7, 11]]
e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
B := [[1, 10], [2, 5], [6, 9]]

Polynomial

x^12+x^11+3*x^10+6*x^9+11*x^8+13*x^7+17*x^6+13*x^5+11*x^4+6*x^3+3*x^2+x+1

Merry Christmas!

Thanks. A very nice Christmas gift!

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
> > Gene,
> >
> > If it's not too much trouble, could you run S(4) X C(4) and also
> > S(4) X S(3)? And give the generators and polynomials, if its not
too
> > much trouble
>
> S(4)XC(3) order 72
>
> Generators
>
> a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> B := [[1, 10], [2, 5], [6, 9]]
>
> Polynomial
>
>
x^12+x^11+3*x^10+8*x^9+14*x^8+19*x^7+24*x^6+19*x^5+14*x^4+8*x^3+3*x^2+
x+1
>
>
> S(4)XS(3) order 144
>
> Generators
>
> a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> b := [[1, 5], [2, 10], [4, 8], [7, 11]]
> e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> B := [[1, 10], [2, 5], [6, 9]]
>
> Polynomial
>
>
x^12+x^11+3*x^10+6*x^9+11*x^8+13*x^7+17*x^6+13*x^5+11*x^4+6*x^3+3*x^2+
x+1
>
>
> Merry Christmas!