# The Universe of Disco

Sun, 30 Apr 2023

Katara is taking a discrete mathematics course this semester, and was asked to prove that if !!D_n!! is the number of derangements of !!n!! objects, then !!D!! satisfies the recurrence $$D_n = (n-1)D_{n-1} + (n-1)D_{n-2}\tag{\star}$$ for !!n\gt 2!!. (A derangement is a permutation with no fixed points.) This is OEIS A000166. (We also have !!D_0=1, D_1=0!!).

She was struggling with this task, and I learned why: She was still thinking of permutations like this:

\begin{align} 1 & \to 2 \\ 2 & \to 4 \\ 3 & \to 5 \\ 4 & \to 1 \\ 5 & \to 3 \end{align}

when she should have been thinking of them like this:

$$(1\; 2\; 4)(3\; 5).$$

“Didn't they teach you about cycle notation?” I asked. Apparently not. (I'm not sure it's a very good class, but it's also possible that this was discussed in one of the lectures that she did not attend.)

Anyway the proof is not difficult, if you think of the cycle structure of the permutations. Let !!p!! be a derangement of size !!n>1!! and consider the location of !!n!! in its cycle structure. The element !!n!! is in some cycle, say one with length !!c!!. !!p!! has no fixed points, so !!c!! is at least !!2!!. If !!c!! is at least !!3!!, we can drop !!n!! from its cycle and still have a derangement of size !!n-1!!:

$$(n\; \ldots)c_2c_3\cdots \Rightarrow (\ldots)c_2c_3\cdots.\tag1$$

If !!n!! is in a cycle of length !!c=2!! and we drop it, then we no longer have a derangement, because we have a permutation which has the fixed point !!p(n)!!. Dropping this also does give us another derangement, of size !!n-2!!:

$$(n\; p(n))c_2c_3\cdots \Rightarrow \underbrace{\color{darkred}{(p(n))}}_{\text{fixed point!}} c_2c_3\cdots \Rightarrow c_2c_3\cdots.\tag2$$

In the former case !!(1)!! there are !!n-1!! positions at which we can insert !!n!! back into the smaller derangement, and each one produces a different permutation, so there are !!(n-1)D_{n-1}!! derangements we can obtain in this way. In the latter case !!(2)!! there are !!n-1!! choices for !!p(n)!!, the element that will share the !!2!!-cycle with !!n!!, so !!(n-1)D_{n-2}!! derangements we can obtain in this way. The total is the sum of these, given by !!(\star)!!.

But if you don't know about permutation cycle structure, you are missing the basic tool that lets you find this argument. To find it, you will probably end up reinventing the idea of cycle structure.

And if you don't know about the representation of permutations in cycle notation, you will have trouble thinking about the cycle structure.

Katara asked me about this in the parking lot outside her dorm just as I was about to drop her off at night, so I was not able to intelligently explain why the cycle structure is nearly always the right way to think about permutations. “It's just better,” I asserted baldly and unconvincingly. After I got home, I tried to think: why is the cycle notation better? It is better, but this is an empirical observation. What is the theoretical cause of its being better?

Most obviously, it's more compact, but this is not the main reason.

A permutation is a bijective function on some carrier set of labels: $$p: S\to S$$ so the only things you need to know about it are: what is !!S!!, and what is !!p!!? In some applications, !!S!! matters. For example, we might want to think of permutations of the English alphabet that leave the set !!\{A, E, I , O, U\}!! fixed.

But typically !!S!! is unimportant, as it is in the derangements problem. In counting derangements, we don't care whether the things being deranged are letters or numbers or hats or fruits or whatever; the identity of !!S!! is completely irrelevant and we took it to be !!\{1, 2, 3, \ldots, n\}!! only because the positive integers are a plentiful supply of items with short names.

And:

If you aren't interested in the labels, the cycle structure of a permutation is the only remaining property it has, because two permutations with the same cycle structure are identical under some change of labels.

Or in jargon, if you are thinking of the permutations as elements of the abstract symmetric group !!S_n!!, relabeling is a conjugation of !!S_n!!, and the cycle classes are exactly the conjugacy classes.

So the cycle structure is exactly the structure of a permutation that remains if you ignore the actual labels, and the cycle notation brings that structure to the foreground.