Anyways, The Unrelated odd thing is: this theorem is called Frobenius, presumably because Frobenius wrote something about it, but what I most think of when I hear “Frobenius” are A) the finite field automorphisms and B) character theory. So I’m wondering if either or both tools together don’t produce something helpful.

]]>Specifically, b(d)=M_d, B(d)=N_d,

a(d)=aperiodic(|G|/(n/d), d)

A(d)=|G|/(n/d) choose d

Then Burnside’s lemma gives you the congruence on (b*A)(n), but this is equal to (B*a)(n), which is what you want.

Pedagogically, I don’t recommend this amount of abstraction. As I said above, this is just expanding the binomial coefficient in terms of the aperiodic terms, regrouping, and combining M into N. And that’s how I first found it (actually, going the other direction). But I wrote it out in the formalism to convince myself that it really is that simple, that I haven’t made any convenient errors. Also, I was worried about comments #5-6, but the convolution formalism takes care of that coincidence, so they aren’t relevant.

PS – necklaces are a somewhat natural fit with cyclic group actions. Just interlace all the orbits. It’s probably a sign of a more direct argument.

]]>Let be the coefficients such that is equivalent to . So we get . This gives

. Switching summation . It seems to me that, for the necklace argument, I want to know that the inner sum is the same as . So I want , which is the surprising fact.

Or did I do this wrong/miss a trick?

]]>Probably you could also first show it for prime-powers, then stitch these together using inclusion-exclusion. (At that point however, you have something equivalent to, though possibly even more complicated than, Möbius inversion.)

I can definitely see why inclusion-exclusion might come up. If you’re counting the solutions to g^6 = 1, then g^2 = 1 and g^3 = 1 seem very likely to come up as subproblems.

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