Why don't any pentachords have transpositional symmetry?

Question

Other than in the trivial case of T₀, none of the thirty-eight pentachords have transpositional symmetry. This makes them unique among all 12-tone sets (well, OK, septachords are the same way, but they're just the complements of the pentachords). I'm having trouble expressing the question: is there a specific mathematical reason for this lack of symmetry? In a way, the answer is obviously yes, since it's true, but I guess what I mean is: if you had to provide a proof that no 12-tone-based set of five notes could ever have transpositional symmetry, what would it be?

Edited to add:

There are several fantastic answers to this question. Anyone coming to the question should really look at several answers, not just the one I accepted. The accepted answer is great, and has the most rigorous proof, but the one by Aaron is probably the most intuitive. The one by cmaster provides some very helpful details, as does the fantastic look into non-12-tone situations by leftaroundabout. These answers in combination will really help any readers get a deeper understanding of the question. Thanks all!

Answer 1

tl;dr skip to the part marked with bold font.

First of all thanks to @Aaron and @PatMuchmore for helping me to understand the formalism presented in learnmusictheory.com.

• Let's consider a system with N pitch classes (you can take N = 12).

• A set x of s elements can be presented as: x = [x₀, x₁, ..., x_s-1].

• In prime form x₀ = 0 (the first note is C).

• A transposition by interval t is the following operation: T_t(x_i) = (x_i + t) mod N.

• A set has transpositional symmetry if there is such pair (or pairs) of t and k so that for each element of set T_t(x_i) = x_{(i+k) mod s} (each set element is transformed into another element of the same set).

• Combining these two formula we get (x_i + t) mod N = x_{(i+k) mod s}. In particular for i = 0 this simplifies to (x₀ + t) = x_k.

• This means that a set with transpositional symmetry w.r.t interval t must contain note t and all its unique multiples. In other words it has a subset in the following form [0, t, 2t, ..., (n-1)·t], where n·t = N

• In addition, in between these notes, there might be (m–1) < t additional notes a_i in form of 0 < a₁ < ...< a_m–1 < t, which repeat by interval t.

• A set symmetric w.r.t. transposition by interval t must therefore have a form:

An example of such set for N=12 and t=4 and m = 3 might be:

[ C,  C#, D           [ 0, 1, 2           [   0,     1,     2   
  E,  F,  Gb     =      4, 5, 6      =        4,   4+1,   4+2   
  G#, A,  Bb ]          8, 9, 10 ]          2·4, 2·4+1, 2·4+2 ]

here a₁ = 1, a₂ = 2. As a_i must be between 0 and t, we can build another combinations, e.g. for a₁ = 1, a₂ = 3:

[ C,  C#, D#          [ 0, 1, 3           [   0,     1,     3   
  E,  F,  G      =      4, 5, 7      =        4,   4+1,   4+3   
  G#, A,  B  ]          8, 9, 11 ]          2·4, 2·4+1, 2·4+3 ]

The number of notes in a set with transpositional symmetry is s = n·m, which must fulfill the following conditions:

• n is a divisor of N (as n·t = N) (note neither t nor n must be prime divisors)
• 1 ≤ m ≤ N/n

As you can see, for N = 12 pitch classes and set with s = n·m = 5 notes, the only pair of n and m which fulfills these condition is n = 1 and m = 5, which makes t = N/n = 12, transposition by octave (or unison), a trivial case.

perhaps the math formalism presented here could be improved, but I hope this is sufficient for a music forum ;)

Answer 2

Simple. 5 does not have any common divisors with 12.

For a chord to have transpositional symmetry, the length of the transposition interval must divide 12, because repeated application of the symmetry needs to return to the original position after some iterations. The transpositionally symmetric chord may subdivide the transposition interval in any way it likes, but the pattern of these subdivision intervals must repeat after a full transposition interval.

For an example, consider a transposition interval of 4 half tone steps (major third): You can choose to take, say C, C#, D# into the chord, but then also E, F and G must be part of it (one transposition up), as must be G#, A and B (two transpositions up). Since the pattern in each transposition interval is the same, the number of notes is a multiple of the number of repetitions.

Now, you can divide 12 like this:

2 parts: tritone transposition
    Chords have 2, 4, 6, 8, 10 or 12 notes

3 parts: major third transposition
    Chords have 3, 6, 9 or 12 notes

4 parts: minor third transposition
    Chords have 4, 8, or 12 notes

6 parts: major second transposition
    Chords have 6 or 12 notes

Thus, the complete list of chord sizes that allow for transpositionally symmetric chords is 2, 3, 4, 6, 8, 9, 10, 12. As you see, all these numbers share at least one divisor with 12, namely the count of transposition intervals that fit into the octave.

Answer 3

The 12-tone octave can be partitioned in two basic, non-intersecting, equal divisions: the tritone division [0, 6], and the major-third division [0, 4, 8]. Past that, any symmetric subset will contain at least one of those basic subdivisions. For example, the 4-division of minor thirds [0, 3, 6, 9] contains as a subset the 2-division, as does [0, 1, 6, 7]. The 6-division contains both the 2- and 3-divisions [0, 2, 4, 6, 8, T], as does [0, 1, 4, 5, 8, 9]. We need not go beyond six pitches, since anything beyond that is just the complement of a less-than-six pitch partition.

The part that needs proof, but which I'm claiming as sufficiently clear, is that any symmetric partition of the the 12-tone octave must be symmetric under T₄ and/or T₆.

This isn't possible for a pentachord. For example, suppose a pentachord contains [0, 6]. Then there must be two pitches between 0 and 6 and only 1 between 6 and 0 (the reverse cases are just T₆ of this), making it non-symmetrical under T₆. Similarly a pentachord containing [0, 4, 8], allowing only two additional pitches, would leave one of the major-third subdivisions without an intervening pitch, making it non-symmetrical under T₄. Thus no pentachord can be symmetrical under either of the "required" transpositional levels.

Answer 4

It does indeed boil down to having no common divisors, as explained by cmaster. Here's for all n-EDOs up to 25 the scale-cardinalities i highlighted where there are no transpositional-symmetric ones, together with gcd(n,i):

1-edo

1(1)

2-edo

1(1) 2(2)

3-edo

1(1) 2(1) 3(3)

4-edo

1(1) 2(2) 3(1) 4(4)

5-edo

1(1) 2(1) 3(1) 4(1) 5(5)

6-edo

1(1) 2(2) 3(3) 4(2) 5(1) 6(6)

7-edo

1(1) 2(1) 3(1) 4(1) 5(1) 6(1) 7(7)

8-edo

1(1) 2(2) 3(1) 4(4) 5(1) 6(2) 7(1) 8(8)

9-edo

1(1) 2(1) 3(3) 4(1) 5(1) 6(3) 7(1) 8(1) 9(9)

10-edo

1(1) 2(2) 3(1) 4(2) 5(5) 6(2) 7(1) 8(2) 9(1) 10(10)

11-edo

1(1) 2(1) 3(1) 4(1) 5(1) 6(1) 7(1) 8(1) 9(1) 10(1) 11(11)

12-edo

1(1) 2(2) 3(3) 4(4) 5(1) 6(6) 7(1) 8(4) 9(3) 10(2) 11(1) 12(12)

13-edo

1(1) 2(1) 3(1) 4(1) 5(1) 6(1) 7(1) 8(1) 9(1) 10(1) 11(1) 12(1) 13(13)

14-edo

1(1) 2(2) 3(1) 4(2) 5(1) 6(2) 7(7) 8(2) 9(1) 10(2) 11(1) 12(2) 13(1) 14(14)

15-edo

1(1) 2(1) 3(3) 4(1) 5(5) 6(3) 7(1) 8(1) 9(3) 10(5) 11(1) 12(3) 13(1) 14(1) 15(15)

16-edo

1(1) 2(2) 3(1) 4(4) 5(1) 6(2) 7(1) 8(8) 9(1) 10(2) 11(1) 12(4) 13(1) 14(2) 15(1) 16(16)

17-edo

1(1) 2(1) 3(1) 4(1) 5(1) 6(1) 7(1) 8(1) 9(1) 10(1) 11(1) 12(1) 13(1) 14(1) 15(1) 16(1) 17(17)

18-edo

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

19-edo

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

20-edo

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

21-edo

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

22-edo

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

23-edo

1(1) 2(1) 3(1) 4(1) 5(1) 6(1) 7(1) 8(1) 9(1) 10(1) 11(1) 12(1) 13(1) 14(1) 15(1) 16(1) 17(1) 18(1) 19(1) 20(1) 21(1) 22(1) 23(23)

24-edo

1(1) 2(2) 3(3) 4(4) 5(1) 6(6) 7(1) 8(8) 9(3) 10(2) 11(1) 12(12) 13(1) 14(2) 15(3) 16(8) 17(1) 18(6) 19(1) 20(4) 21(3) 22(2) 23(1) 24(24)

25-edo

1(1) 2(1) 3(1) 4(1) 5(5) 6(1) 7(1) 8(1) 9(1) 10(5) 11(1) 12(1) 13(1) 14(1) 15(5) 16(1) 17(1) 18(1) 19(1) 20(5) 21(1) 22(1) 23(1) 24(1) 25(25)

Note that exactly those where the GCD is 1 are highlighted.

Source code (in Haskell):

import Data.List

transpositions :: [a] -> [[a]]
transpositions l = map (take ll) . take ll $ iterate tail $ cycle l
 where ll = length l

isTpSym :: Eq a => [a] -> Bool
isTpSym scl = any (==scl) . tail $ transpositions scl

type Scale = [Bool]

allScales :: Int -> [Scale]
allScales 0 = [[]]
allScales n = [False, True] >>= \h -> (h:)<$>allScales (n-1)

sclsInfo :: Int -> String
sclsInfo n = show n++"-edo\n---\n"
           ++ intercalate " "
             [ (if i`elem`tscs then id else \s->"**"++s++"**")
                 $ show i++"("++show(gcd i n)++")"
             | i <- [1..n] ]
           ++"\n"
 where tscs = map head . group . sort
             $ [ length $ filter id scl
               | scl <- allScales n, isTpSym scl ]

main :: IO ()
main = mapM_ putStrLn $ sclsInfo<$>[1..25]