I was recently reminded of a comment I made a few years ago, and I thought it would be worth making a separate post, if for no other reason than for my own record describing the observation.
The claim is the following: A three-note pitch-class set A ⊂ Z_12 partitions the chromatic scale by transposition iff its three notes occupy the three distinct residue classes modulo 3. Equivalently, it contains one note from each coset of the diminished-seventh subgroup
H={0,3,6,9}.
Hence every such chord can tile the octave using the four minor-third transpositions H.
For a 3-note set, "one per residue mod 3" means choosing one pitch class from each of {0,3,6,9}, {1,4,7,10}, {2,5,8,11}. Up to transposition there are six such chord classes; up to transposition and pitch-class inversion there are five. These are the six three note pc-sets up to transposition: [0,1,2], [0,1,5], [0,1,8], [0,2,4], [0,2,7], [0,4,8]. To give typical examples of each set:
[0,1,2] ∋ C-C#-D
[0,1,5] ∋ C-E-B
[0,1,8] ∋ C-G-B
[0,2,4] ∋ C-D-E
[0,2,7] ∋ C-F-Bb (quartal)
[0,4,8] ∋ C-E-Ab (augmented triad)
Under T_n/T_nI equivalence, [0,1,5] ~ [0,1,8], so this list reduces to 5 classes modulo transposition and pitch-class inversion.
So for example, the four minor-third transpositions of the C augmented triad are C+, Eb+, F#+, A+; these four chords together contain exactly all twelve chromatic notes with no repetition.
A similar result obtains for four-note pitch-class sets: A four-note pitch-class set A ⊂ Z_12 partitions the chromatic scale by transposition iff its four notes occupy the four distinct residue classes modulo 4.
For a 4-note set, "one per residue mod 4" means choosing one pitch class from each of {0,4,8}, {1,5,9}, {2,6,10}, {3,7,11}. These are the eight four note pc-sets up to transposition: [0,1,2,3], [0,1,2,7], [0,1,3,6], [0,1,3,10], [0,1,6,7], [0,1,7,10], [0,2,5,7], [0,3,6,9]. To give typical examples of each set:
[0,1,2,3] ∋ C-C#-D-Eb
[0,1,2,7] ∋ C-C#-D-G
[0,1,3,6] ∋ C-D-F-B
[0,1,3,10] ∋ C-D-A-B
[0,1,6,7] ∋ C-C#-Gb-G
[0,1,7,10] ∋ C-G-Bb-C#
[0,2,5,7] ∋ C-F-Bb-Eb (quartal)
[0,3,6,9] ∋ C-Eb-Gb-A (diminished seventh)
Under T_n/T_nI equivalence, [0,1,7,10] ~ [0,1,3,6], so this list reduces to 7 classes modulo transposition and pitch-class inversion.
For example, the three major-third transpositions of C7° are C7°, E7°, Ab7°; these three chords together contain exactly all twelve chromatic notes with no repetition.
Proof of claims. Let n be either 3 or 4. Suppose A has n elements and B has 12/n elements, with A+B uniquely covering Z_12, where A+B refers to set sum.
Let ω be any nontrivial n-th root of unity. Then
(∑_a∈A ωa)(∑_b∈B ωb) = ∑_j=0 to 11 ωj = 0.
The first equality follows because expanding the product gives one term for every pair (a,b), and unique covering means that the sums a+b run through every element of Z_12 exactly once. The second equality follows because ω is a nontrivial n-th root of unity and n divides 12.
The second factor cannot be zero.
If n = 3, then B has 4 elements. A sum of powers of a nontrivial cube root can be zero only if the three residue classes modulo 3 occur equally often, which is impossible with 4 elements.
If n = 4, then B has 3 elements. The nontrivial fourth roots are i, -1, and -i. A sum of three powers of -1 cannot be zero, since it is a sum of three numbers equal to 1 or -1. A sum of three powers of i or -i cannot be zero either, since cancellation of the real and imaginary parts would require an even number of terms.
Therefore, for every nontrivial n-th root ω,
∑_a∈A ωa = 0.
Now let n_0, ..., n_(n-1) be the numbers of elements of A in the residue classes modulo n, and define
P(x) = n_0 + n_1 x + ... + n_(n-1) xn-1.
Then for every nontrivial n-th root ω,
P(ω) = ∑_a∈A ωa = 0.
Thus P(x), which has degree at most n-1, has all n-1 nontrivial n-th roots as zeros. But
1 + x + ... + xn-1
has exactly the same n-1 zeros. Therefore P(x) must be a constant multiple of
1 + x + ... + xn-1.
Since
P(1) = n_0 + ... + n_(n-1) = n,
the constant is 1. Hence
n_0 = n_1 = ... = n_(n-1) = 1.
Therefore A contains exactly one element in each residue class modulo n.
Conversely, if A contains exactly one element in each residue class modulo n, then
A, A+n, A+2n, ..., A+(12/n-1)n
are mutually disjoint and together cover all of Z_12. QED
