Group symmetries in 22-tET

(For Gene):

Do you think there would be any value to running the Polya polynomials
(with permutations) for 22-tET?

Of course, there would only be:

D22
C2 * C11
C2 * D11
C2 * S11

Any others?

I've already running the "other" Polya method, which is only good
for Hendecachords in 22-et.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> (For Gene):
>
> Do you think there would be any value to running the Polya
> polynomials (with permutations) for 22-tET?
>
> Of course, there would only be:
>
> D22
> C2 * C11
> C2 * D11
> C2 * S11
>
> Any others?
>

What I am interested in (I am planning to do it, when I get the time),
would be

GL(1,Z22) - the general linear group, i.e. the multiplicative group of
integers that are invertible modulo 22

and

C22 x GL(1,Z22) - i.e. the general affine group.

> I've already running the "other" Polya method, which is only good
> for Hendecachords in 22-et.
>

What's the "other" Polya method?
--
Hans Straub

--- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
> >
> > (For Gene):
> >
> > Do you think there would be any value to running the Polya
> > polynomials (with permutations) for 22-tET?
> >
> > Of course, there would only be:
> >
> > D22
> > C2 * C11
> > C2 * D11
> > C2 * S11
> >
> > Any others?
> >
>
> What I am interested in (I am planning to do it, when I get the
time),
> would be
>
> GL(1,Z22) - the general linear group, i.e. the multiplicative group
of
> integers that are invertible modulo 22
>
> and
>
> C22 x GL(1,Z22) - i.e. the general affine group.
>
> > I've already running the "other" Polya method, which is only good
> > for Hendecachords in 22-et.

(Not true, it can be used for all sets, plus I think I can figure out
how to apply to any subset class) Just not as slick as the Polya
method Gene et. al. uses)
> >
>
> What's the "other" Polya method?

It's the one where you count all the sets in an et or just one class,
like hexachords in 12-et (80). Here's the formula for that:

a(n) = sum {d|n} (phi(n/d)*C(2d,d))/(2n), n>0.

To count all the sets, such as 352 sets in 12-et:

a(n) = (1/n)*Sum_{ d divides n } phi(d)*2^(n/d).

Paul Hj

>

--- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...> wrote:

> GL(1,Z22) - the general linear group, i.e. the multiplicative group of
> integers that are invertible modulo 22

This would most often be called (Z/22Z)*. You probably know this
already, but I'll mention it is a cyclic group of order 10; the
generator can be taken as 7.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
wrote:
>
> > GL(1,Z22) - the general linear group, i.e. the multiplicative group
of
> > integers that are invertible modulo 22
>
> This would most often be called (Z/22Z)*. You probably know this
> already, but I'll mention it is a cyclic group of order 10; the
> generator can be taken as 7.
>
Can you expand on this a little? Only 10 elements? Also, tell me
if S11*C2, D11*C2 are completely ridiculous (I take it that C11XC2 and
D22 probably cover everything that needs to be addressed)

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> > This would most often be called (Z/22Z)*. You probably know this
> > already, but I'll mention it is a cyclic group of order 10; the
> > generator can be taken as 7.
> >
> Can you expand on this a little? Only 10 elements?

The multiplicative group of numbers mod 22 can be determined from the
fact that C22 = C2 x C11. C11, considered as Z/11Z, has a
mulitpliciative group (group if units) with 10 = 11-1 elements and is
cyclic, since 11 is prime. In the case of 2, p-1 is 1 and the
multiplicative group is trivial. Hence, the group of units of numbers
mod 22 is cyclic of order 10. You can check that 7 generates it.

Also, tell me
> if S11*C2, D11*C2 are completely ridiculous (I take it that C11XC2 and
> D22 probably cover everything that needs to be addressed)

They look awfully large to me.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > > This would most often be called (Z/22Z)*. You probably know this
> > > already, but I'll mention it is a cyclic group of order 10; the
> > > generator can be taken as 7.
> > >
> > Can you expand on this a little? Only 10 elements?
>
> The multiplicative group of numbers mod 22 can be determined from
the
> fact that C22 = C2 x C11. C11, considered as Z/11Z, has a
> mulitpliciative group (group if units) with 10 = 11-1 elements and
is
> cyclic, since 11 is prime. In the case of 2, p-1 is 1 and the
> multiplicative group is trivial. Hence, the group of units of
numbers
> mod 22 is cyclic of order 10. You can check that 7 generates it.

I see. Sorry to be dense, but how does 7 generate it? Do you subtract
1 from 11 and from 2 because of the number less than x relatively
prime to x? If I am on the right track this is a Multiplicative
Modulo Group right?

> Also, tell me
> > if S11*C2, D11*C2 are completely ridiculous (I take it that
C11XC2 and
> > D22 probably cover everything that needs to be addressed)
>
> They look awfully large to me.

Jon Wild and I are working with sets with huge numbers. Cygwin
has been great to work with even though I am not as happy with Octave
itself.
>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@a...> wrote:
> >
> > > > This would most often be called (Z/22Z)*. You probably know
this
> > > > already, but I'll mention it is a cyclic group of order 10;
the
> > > > generator can be taken as 7.
> > > >
> > > Can you expand on this a little? Only 10 elements?
> >
> > The multiplicative group of numbers mod 22 can be determined from
> the
> > fact that C22 = C2 x C11. C11, considered as Z/11Z, has a
> > mulitpliciative group (group if units) with 10 = 11-1 elements
and
> is
> > cyclic, since 11 is prime. In the case of 2, p-1 is 1 and the
> > multiplicative group is trivial. Hence, the group of units of
> numbers
> > mod 22 is cyclic of order 10. You can check that 7 generates it.
>
> I see. Sorry to be dense, but how does 7 generate it? Do you
subtract
> 1 from 11 and from 2 because of the number less than x relatively
> prime to x? If I am on the right track this is a Multiplicative
> Modulo Group right?

I got it - it's because 7 generates 1, 3, 5, 7, 9, 13, 15, 17, 19, 21
(These are the numbers relatively prime to 22)

>
> > Also, tell me
> > > if S11*C2, D11*C2 are completely ridiculous (I take it that
> C11XC2 and
> > > D22 probably cover everything that needs to be addressed)
> >
> > They look awfully large to me.
>
> Jon Wild and I are working with sets with huge numbers. Cygwin
> has been great to work with even though I am not as happy with
Octave
> itself.
> >
>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> I see. Sorry to be dense, but how does 7 generate it? Do you subtract
> 1 from 11 and from 2 because of the number less than x relatively
> prime to x? If I am on the right track this is a Multiplicative
> Modulo Group right?

If you take 7^n mod 22, for n from 0 to 9, you get all of the numbers
from 0 to 21 relatively prime to 22. Then 7^10 mod 22 is 1 again.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > I see. Sorry to be dense, but how does 7 generate it? Do you
subtract
> > 1 from 11 and from 2 because of the number less than x relatively
> > prime to x? If I am on the right track this is a Multiplicative
> > Modulo Group right?
>
> If you take 7^n mod 22, for n from 0 to 9, you get all of the numbers
> from 0 to 21 relatively prime to 22. Then 7^10 mod 22 is 1 again.
>
Right. At first I was doing 7*n, which is clearly incorrect.
Now I can see where C11XC2 etc. are valuable in tuning theory,
but not sure what use the multiplication modulo group would be.
Does it have any immediate applications?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> Right. At first I was doing 7*n, which is clearly incorrect.
> Now I can see where C11XC2 etc. are valuable in tuning theory,
> but not sure what use the multiplication modulo group would be.
> Does it have any immediate applications?

It's applicable to the issue you introduced, namely permutation
groups, and Aff(n), the affine group on n elements, in particular.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > Right. At first I was doing 7*n, which is clearly incorrect.
> > Now I can see where C11XC2 etc. are valuable in tuning theory,
> > but not sure what use the multiplication modulo group would be.
> > Does it have any immediate applications?
>
> It's applicable to the issue you introduced, namely permutation
> groups, and Aff(n), the affine group on n elements, in particular.
>
Okay. So how would one use 1,3,5,7,9,13,15,17,19,21 in 22 et, for
example? (MMG)

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> Okay. So how would one use 1,3,5,7,9,13,15,17,19,21 in 22 et, for
> example? (MMG)

Each of these introduces a permutation by sending x mod 22 to a*x mod
22. In terms of generators, it sends generators for one temperament
supported by 22 to another.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > Okay. So how would one use 1,3,5,7,9,13,15,17,19,21 in 22 et, for
> > example? (MMG)
>
> Each of these introduces a permutation by sending x mod 22 to a*x mod
> 22. In terms of generators, it sends generators for one temperament
> supported by 22 to another.
>
Interesting. So these are values for a? Let's try it 5 mod 22=5, 7*5
mod 22= 35 mod 22=13?

In terms of generators, I take it 7 will determine these 10 values.
Then do you fill in the other 12 values somehow?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> In terms of generators, I take it 7 will determine these 10 values.
> Then do you fill in the other 12 values somehow?

The other twelve values don't work, since they send 22 notes to the
octave to something less.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@a...> wrote:
> >
> > > Okay. So how would one use 1,3,5,7,9,13,15,17,19,21 in 22 et, for
> > > example? (MMG)
> >
> > Each of these introduces a permutation by sending x mod 22 to a*x
mod
> > 22.

Since you seem to be at your desk, could you show this? (What's a?)
Thanks

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> Since you seem to be at your desk, could you show this? (What's a?)
> Thanks

The "a" is relatively prime to 22. If a=3, for example, then
3*n sends 0-21 in order to:

[0 3 6 9 12 15 18 21 2 5 8 11 14 17 20 1 4 7 10 13 16 19]

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
>
> > Since you seem to be at your desk, could you show this? (What's a?)
> > Thanks
>
> The "a" is relatively prime to 22. If a=3, for example, then
> 3*n sends 0-21 in order to:
>
> [0 3 6 9 12 15 18 21 2 5 8 11 14 17 20 1 4 7 10 13 16 19]
>

primitive generators. I read too much into this, but thanks.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:

> primitive generators. I read too much into this, but thanks.

There is more to be said, in that sending x to a*x+b mod 22, the
affine group, is a semidirect product of two cyclic groups, and the
10-cyclic part is generated by powers of 7.

Hans Straub wrote:

> What I am interested in (I am planning to do it, when I get the
> time), would be GL(1,Z22) - the general linear group, i.e. the
> multiplicative group ofintegers that are invertible modulo 22
> and C22 x GL(1,Z22) - i.e. the general affine group.
>
> Hans Straub

Hans or Gene or anyone who knows,

Is this Aff(22)? I am still learning about GL(1,Z22) on sci.math,
but would this be the same?

GAP code:

G:= Group( [ (1,3,9,5,15)(2,6,18,10,8)(4,12,14,20,16)(7,21,19,13,17),

(1,5,3,15,9)(2,10,6,8,18)(4,20,12,16,14)(7,13,21,17,19),

(1,9,15,3,5)(2,18,8,6,10)(4,14,16,12,20)(7,19,17,21,13),

(1,7,5,13,3,21,15,17,9,19)(2,14,10,4,6,20,8,12,18,16),

(1,13,15,19,5,21,9,7,3,17)(2,4,8,16,10,20,18,14,6,12),

(1,15,5,9,3)(2,8,10,18,6)(4,16,20,14,12)(7,17,13,19,21),

(1,17,3,7,9,21,5,19,15,13)(2,12,6,14,18,20,10,16,8,4),

(1,19,9,17,15,21,3,13,5,7)(2,16,18,12,8,20,6,4,10,14),

(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22),

(1,21)(2,20)(3,19)(4,18)(5,17)(6,16)(7,15)(8,14)(9,13)(10,12) ] );

pol :=CycleIndex(G);

1/220*x_1^22+1/20*x_1^2*x_2^10+3/55*x_2^11+1/5*x_1^2*x_5^4+1/5*x_1^2*x
_10^2+2/5*x_2*x_10^2+1/22*x_11^2+1/22*x_22

x_1:=Indeterminate(Integers,1);
x_2:=Indeterminate(Integers,2);
x_5:=Indeterminate(Integers,5);
x_10:=Indeterminate(Integers,10);
x_11:=Indeterminate(Integers,11);
x_22:=Indeterminate(Integers,22);

Ans:=Value(pol,[x_1,x_2,x_5,x_10,x_11,x_22],
x_1+1,x_1^2+1,x_1^5+1,x_1^10+1,x_1^11+1,x_1^22+1]);

x_1^22+x_1^21+3*x_1^20+8*x_1^19+39*x_1^18+125*x_1^17+358*x_1^16+788*x_
1^15+1488*x_1^14+2282*x_1^13+2990*x_1^12+3235*x_1^11+2990*x_1^10+2282*
x_1^9+1488*x_1^8+788*x_1^7+358*x_1^6+125*x_1^5+39*x_1^4+8*x_1^3+3*x_1^
2+x_1+1

It has order 220, so is this C22 X (Z/Z22)*?

Thanks,

PGH
Thanks,

PGH

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<phjelmstad@...> wrote:
>
> Hans or Gene or anyone who knows,
>
> Is this Aff(22)? I am still learning about GL(1,Z22) on sci.math,
> but would this be the same?
>
> GAP code:
>
> G:= Group( [ (1,3,9,5,15)(2,6,18,10,8)(4,12,14,20,16)
(7,21,19,13,17),
>
> (1,5,3,15,9)(2,10,6,8,18)(4,20,12,16,14)(7,13,21,17,19),
>
> (1,9,15,3,5)(2,18,8,6,10)(4,14,16,12,20)(7,19,17,21,13),
>
> (1,7,5,13,3,21,15,17,9,19)(2,14,10,4,6,20,8,12,18,16),
>
> (1,13,15,19,5,21,9,7,3,17)(2,4,8,16,10,20,18,14,6,12),
>
> (1,15,5,9,3)(2,8,10,18,6)(4,16,20,14,12)(7,17,13,19,21),
>
> (1,17,3,7,9,21,5,19,15,13)(2,12,6,14,18,20,10,16,8,4),
>
> (1,19,9,17,15,21,3,13,5,7)(2,16,18,12,8,20,6,4,10,14),
>
> (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22),
>
> (1,21)(2,20)(3,19)(4,18)(5,17)(6,16)(7,15)(8,14)(9,13)(10,12) ] );
>
> pol :=CycleIndex(G);
>
>
1/220*x_1^22+1/20*x_1^2*x_2^10+3/55*x_2^11+1/5*x_1^2*x_5^4+1/5*x_1^2*x
> _10^2+2/5*x_2*x_10^2+1/22*x_11^2+1/22*x_22
>
> x_1:=Indeterminate(Integers,1);
> x_2:=Indeterminate(Integers,2);
> x_5:=Indeterminate(Integers,5);
> x_10:=Indeterminate(Integers,10);
> x_11:=Indeterminate(Integers,11);
> x_22:=Indeterminate(Integers,22);
>
> Ans:=Value(pol,[x_1,x_2,x_5,x_10,x_11,x_22],
> x_1+1,x_1^2+1,x_1^5+1,x_1^10+1,x_1^11+1,x_1^22+1]);
>
>
x_1^22+x_1^21+3*x_1^20+8*x_1^19+39*x_1^18+125*x_1^17+358*x_1^16+788*x_
>
1^15+1488*x_1^14+2282*x_1^13+2990*x_1^12+3235*x_1^11+2990*x_1^10+2282*
>
x_1^9+1488*x_1^8+788*x_1^7+358*x_1^6+125*x_1^5+39*x_1^4+8*x_1^3+3*x_1^
> 2+x_1+1
>
> It has order 220, so is this C22 X (Z/Z22)*?
>

Yes, that's the affine group of Z22. Thanks for posting this -
especially the GAP code. Quite amazing how easy this goes in GAP! I
gotta spend more time with GAP...
--
Hans Straub

--- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <phjelmstad@> wrote:
> >
> > Hans or Gene or anyone who knows,
> >
> > Is this Aff(22)? I am still learning about GL(1,Z22) on sci.math,
> > but would this be the same?
> >
> > GAP code:
> >
> > G:= Group( [ (1,3,9,5,15)(2,6,18,10,8)(4,12,14,20,16)
> (7,21,19,13,17),
> >
> > (1,5,3,15,9)(2,10,6,8,18)(4,20,12,16,14)(7,13,21,17,19),
> >
> > (1,9,15,3,5)(2,18,8,6,10)(4,14,16,12,20)(7,19,17,21,13),
> >
> > (1,7,5,13,3,21,15,17,9,19)(2,14,10,4,6,20,8,12,18,16),
> >
> > (1,13,15,19,5,21,9,7,3,17)(2,4,8,16,10,20,18,14,6,12),
> >
> > (1,15,5,9,3)(2,8,10,18,6)(4,16,20,14,12)(7,17,13,19,21),
> >
> > (1,17,3,7,9,21,5,19,15,13)(2,12,6,14,18,20,10,16,8,4),
> >
> > (1,19,9,17,15,21,3,13,5,7)(2,16,18,12,8,20,6,4,10,14),
> >
> > (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22),
> >
> > (1,21)(2,20)(3,19)(4,18)(5,17)(6,16)(7,15)(8,14)(9,13)(10,12) ] );
> >
> > pol :=CycleIndex(G);
> >
> >
>
1/220*x_1^22+1/20*x_1^2*x_2^10+3/55*x_2^11+1/5*x_1^2*x_5^4+1/5*x_1^2*x
> > _10^2+2/5*x_2*x_10^2+1/22*x_11^2+1/22*x_22
> >
> > x_1:=Indeterminate(Integers,1);
> > x_2:=Indeterminate(Integers,2);
> > x_5:=Indeterminate(Integers,5);
> > x_10:=Indeterminate(Integers,10);
> > x_11:=Indeterminate(Integers,11);
> > x_22:=Indeterminate(Integers,22);
> >
> > Ans:=Value(pol,[x_1,x_2,x_5,x_10,x_11,x_22],
> > x_1+1,x_1^2+1,x_1^5+1,x_1^10+1,x_1^11+1,x_1^22+1]);
> >
> >
>
x_1^22+x_1^21+3*x_1^20+8*x_1^19+39*x_1^18+125*x_1^17+358*x_1^16+788*x_
> >
>
1^15+1488*x_1^14+2282*x_1^13+2990*x_1^12+3235*x_1^11+2990*x_1^10+2282*
> >
>
x_1^9+1488*x_1^8+788*x_1^7+358*x_1^6+125*x_1^5+39*x_1^4+8*x_1^3+3*x_1^
> > 2+x_1+1
> >
> > It has order 220, so is this C22 X (Z/Z22)*?
> >
>
> Yes, that's the affine group of Z22. Thanks for posting this -
> especially the GAP code. Quite amazing how easy this goes in GAP! I
> gotta spend more time with GAP...
> --
> Hans Straub
>

Thanks! GAP is cool, and the GAP Forum is great for getting ideas
for coding. I just cracked the "5-clustering of Z-relations in 22tET"
based on what I found here in the Affine Group. (Actually, all the
sets 5 cluster cuz of the 5-cycles and 10-cycles here). Instead
of posting it I will upload a paper to Files.

PGH