I found something interesting while studying the circle of fifths.
Define a map $f(x) = 7x$ (mod 12), this models the circle of fifths as a table. The input is the location on the circle of fifths, the output is the note at that location on the circle of fifths.
You could also read the input as the number of sharps or flats in the scale. So $f(2) = 14 \equiv 2$ (mod 12) reads "the scale with two sharps is D" or "two o'clock on the circle of fifths is D". And $f(-1) = -7 \equiv 5$ (mod 12) reads "the scale with one flat is F".
The function f is an involution, because $f(f(x)) = 7(7x) = 49x= (48 + 1)x \equiv x$ (mod 12).
While pairing up the elements, a beautiful relation appeared:
A line between two numbers means those numbers map to each other, no line means the number maps to itself. It's easy to verify this picture, Proof:
Evens are fixed because: $f(2n) = 7(2n) = 14n = (12 + 2)n \equiv 2n$ (mod 12)
Odds map to the opposite side of the circle because: $f(2n + 1) = f(2n) + f(1) = 2n + f(1) = 2n + 7 = (2n + 1) + 6$
My goal is to understand the above picture better, I'm open to general suggestions about how to do this, but I got stuck trying something very specific.
I'm trying to articulate the question "Prove the symmetries of this function are $D_{12}$" I've begun by defining: $i(n) = n$ (mod 12) $r(n) = n + 2$ (mod 12) $s(n) = -n$ (mod 12) These satisfy: $r^6 = s^2 = (rs)^2 = i$ which make them generators of $D_{12}$.
I'm struggling to continue my argument, here is my current attempt:
Now let $D_{12}$ act on $\mathbf{H}(\mathbf{Z}/12\mathbf{Z}, \mathbf{Z}/12\mathbf{Z})$ by conjugation. So for $g \in \mathbf{H}(\mathbf{Z}/12\mathbf{Z}, \mathbf{Z}/12\mathbf{Z})$ and $p \in D_{12}$ we have $p \cdot g = p g p^{-1}$.
Now we can show $D_{12}$ fixes $f$, it follows from $r \cdot f (x) = r f r^{-1} (x) = - f(- x) = - 7(- x) = 7x = f(x)$ and $s \cdot f(x) = s f s^{-1} (x) = f(x - 2) + 2 = f(x) - 2 + 2 = f(x)$
I have a handful of misgivings with my thoughts so far. I don't know what larger group should be thought of as containing $D_{12}$. And $f$ being a fixed point of a group doesn't prove it isn't a fixed point of some larger group.
Right now, my current approach for a fix of this is to prove that the stabilizer subgroup of the permutations with respect to $f$ equals the dihedral group: ${S_{12}}_f = D_{12}$.
Is there a more specific ambient space than just permutations though? Aren't I cheating by applying an arbitrary permutation to a ring isomorphism. And that's what my map is right? Or is it actually a $Z/12Z$ - module isomorphism? I feel very stuck on what I think are some pretty basic concepts.
I have a strong intuition for the symmetries of this circle now, For example from the above it's clear that f having one flat is equivalent to b having five sharps. But I'm struggling translating these intuitions into algebraic proofs.
I would appreciate any help incorporating $D_{12}$ into wherever it belongs.
Best,
Edit: I tried to clarify my definition.
I'm getting suggestions that I either meant, or should be using the function $m(k) = k + 7$ (mod 12). I labeled it "$m$" for modulate. It has the advantage that the inputs and the outputs are both semitones. It's related to my function by:
$f(n) = m^n(0)$
Or "the nth location on the circle of fifths is labeled by the note you get by finding n fifths from C"
A partial answer:
Your clock picture is saying that for $x$ odd then $7x \equiv x+6\bmod 12$.
This is clearly true $\bmod 3$ (for all $x$) so it remains to check $\bmod 4$ where this is clearly true again (for $x$ odd, and it fails for $x$ even).
From the music perspective it is just saying that $2n$ perfect fifts = $2n$ semitones.
I don't see any reason to restrict to $x$ odd and to seek $D_{12}$ symmetries in there. $D_{12}$ is a subgroup of $Aff(\Bbb{Z/12Z})$ and the natural embedding of $D_6$ acts on $\Bbb{2Z/12Z}$ thus on your $6$ points.