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Notice that $$3^4 - 3 = 81 - 3 = 78,$$ which is divisible by $2$, but not by $4$.

In your opinion, what "goes wrong" in this case?

This is a question meant to be answered via combinatorics, and to me, what I was able to note that there may be $(i,j)$ repetitions, such that $3*3 - 3 = 6$, which $6$ is not divisible by $4$.

Can someone explain what goes wrong in terms of the combinatorics?

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2 Answers 2

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The point of Qiaochu Yuan's answer is that firstly that in order to be surprised the $a^p-a$ is not divisible by $p$ when $p$ is composite, you should first have a (combinatorial in your case) reason to find it natural that it does hold when $p$ is a prime number. One such reason is that $a^p-a$ counts words (which I find more convenient to talk about than necklaces, but it is the same thing) of length $p$ over an $a$-element alphabet that do not consist of $p$ identical letters. The reason this number is divisible by $p$ is that cyclic rotation partitions the set of such words into packets (orbits) of size $p$. That this is so requires a bit of thought: if there were any non-trivial cyclic rotation (not necessarily by a single place) that transforms a word into itself, then by iterating that one can obtain the cyclic rotation by a single place, and any word invariant by that rotation clearly has all its letters equal. This amounts to the fact that any non-identity element of a cyclic group of $p$ element generates the entire group.

Now the second point of the Qiaochu Yuan's answer is to analyze the situation when $p$ is composite (I'll henceforth call the number $n$ to avoid confusion). Now it is no longer true that any non-identity element of a cyclic group of $n$ element generates the entire group; notably it can generate a subgroup of any order dividing $n$. Consequently there are words that are invariant under some non-trivial cyclic rotation, but not under all of them, and in particular not under the rotation by a single place. The word RBYRBY is an example, invariant under rotation by $3$ places, but not under rotation by one place. You may very well stop here; it is clear why the argument breaks down, and the numerical examples you gave show that there is no other argument that will make $a^n-a$ divisible by $n$ for all $n$ either.

If you insist on finding something that is counted by a number always divisible by $n$, then you can still take all words that are not invariant under any non-trivial cyclic rotation, as these come in packets (orbits) of size $n$. Call this number $f(n)$, which will be useful later. Now rather than having $f(n)=a^n-a$, you need to exclude for every strict divisor $d$ of $n$ the words that come in packets of size $d$. These are harder to count than words that come in singleton packets (as we saw there were $a$ of them), but can be done by the following reasoning. Such words necessarily consist of a sub-word $s$ of $d$ letters that is repeated $d$ times: $w=s.s\ldots s$. However $s$ cannot just be any word, it must not have any internal cyclic symmetry, since if it did it would belong to a shorter orbit.

But words without any cyclic symmetry is just what we were out to count, except that it was for length $n$ and now we are confronted with length $d<n$. Hence the utility of a general function $f$: it is defined implicitly (by induction) by the relation $$ f(m)=a^m-\sum_{d\text{ strictly divides }m}f(d) $$ for all $m\geq1$. Note that in particular $f(1)=a$ since $1$ has no strict divisors. This equation can be formulated more simply by moving the summation to the other side, which also avoid having to mention "strict": $$ \sum_{d\mid m}f(d)=a^m $$ for all $m\geq1$. The general technique the will solve this is called Möbius inversion. Since you haven't heard about that, here's its secret. Think of a vectors $F(n)=(f(1),f(2),\ldots,f(n))$, the unknown we are after, and $G(n)=(a,a^2,\ldots,a^n)$ which tabulates the right-hand-sides of our last equation. This equation then says $F(n)=\mathbf D(n)\cdot F(n)$, where $\mathbf D(n)$ is the lower triangular matrix with entries $\mathbf D(n)_{i,j}=1$ whenever $j$ divides $i$, and $\mathbf D(n)_{i,j}=0$ otherwise. Actually the equation gives us the result of multiplying row $m$ of of the matrix $\mathbf D(n)$ and the vector $G(n)$, so the full matrix equation takes care of "for all $m$" (upto $n$). Now all we need is the inverse matrix $\mathbf D(n)$, and it can (rather easily) be shown to be given by $$ \mathbf D(n)^{-1}_{i,j} = \begin{cases}(-1)^c & \text{if }i/j\text{ is integer and a product of }c\text{ distinct primes}\\ 0 & \text{otherwise}\\ \end{cases} $$ which is usually written as $$ \mathbf D(n)^{-1}_{i,j} = \begin{cases}\mu(i/j)&\text{if }j\mid i\\ 0 & \text{otherwise}\\ \end{cases} $$ where $\mu$ is the so-called (arithmetic) Möbius function. Now one can write $F(n)=\mathbf D(n)^{-1}\cdot G(n)$, giving $$ f(m)=\sum_{d\mid m}\mu(m/d)a^d $$ which is the formula Qiaochu gave (with $m=n$) for the number of entirely non-symmetric necklacesof length $n$. (Actually he interchanged $n/d$ and $d$, which can be seen to make no difference; personally I prefer to always put $n/d$ in the argument of $\mu$.)

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Thank you for this very thorough explanation, Marc! I confess that I assumed that the OP was already familiar with the combinatorial argument and that I was just resummarizing it to make sure we were on the same page. –  Qiaochu Yuan Jan 24 '12 at 15:56

The combinatorial proof that $p \mid a^p - a$ when $p$ is prime is that $a^p - a$ counts the number of acyclic necklaces of length $p$ with $a$ colors. For any $n$, the cyclic group $\mathbb{Z}/n\mathbb{Z}$ acts by cyclic permutation on the set of necklaces of length $n$ with $a$ colors. When $p$ is prime, we can split up the set of necklaces as follows:

  • The necklaces all of whose colors are the same, of which there are $a$, and
  • The acyclic necklaces, of which there are $a^p - a$.

The action of $\mathbb{Z}/p\mathbb{Z}$ on the first collection is trivial and the action on the second is free because $\mathbb{Z}/p\mathbb{Z}$ has no nontrivial subgroups, so the action on the second collection splits up into $\frac{a^p - a}{p}$ orbits of size $p$ by the orbit-stabilizer theorem.

When $n$ is not prime, other orbit types may appear, since necklaces may have nontrivial cyclic structure such as $RBYRBY$, so there are orbits with nontrivial stabilizers. In this case the correct count of the number of acyclic necklaces (those on which $\mathbb{Z}/n\mathbb{Z}$ acts freely) is given, not by $a^n - a$, but by $$\sum_{d \mid n}\; \mu(d) a^{n/d}$$

where $\mu$ is the Möbius function; the proof is a classical application of Möbius inversion and worth working out as an exercise. So the correct generalization to the non-prime case from the combinatorial point of view is that $$n \mid \sum_{d \mid n}\; \mu(d) a^{n/d}.$$


In the specific case $a = 3, n = 4$, the group $\mathbb{Z}/4\mathbb{Z}$ has $\mathbb{Z}/2\mathbb{Z}$ as its only nontrivial subgroup. Letting the colors be $R, B, Y$, the orbits with full stabilizer look like $RRRR$, the orbits with stabilizer $\mathbb{Z}/2\mathbb{Z}$ look like $RBRB$, and the orbits with trivial stabilizer look like $RBYB$. There are $3$ of the first kind and $$\frac{\mu(1) 3^4 + \mu(2) 3^2 + \mu(4) 3^1}{4} = \frac{3^4 - 3^2}{4} = 18$$

of the third kind, so $3$ of the second kind.

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Can you explain it without using the Mobius function? Our class has never studied that yet, and its not in the book. Thanks –  Buddy Holly Jan 24 '12 at 2:46
    
Also, can you edit your answer to incorporate the Orbit Stabilizer Theorem? Thanks –  Buddy Holly Jan 24 '12 at 2:51
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@Buddy: I have never understood requests of this kind. The Mobius function naturally appears in the answer, so you are asking me not to explain part of the answer. You can just ignore that part of the answer and work out the orbit structure for specific choices of $a$ and $n$ if you want, but I don't see the point of editing out relevant information. –  Qiaochu Yuan Jan 24 '12 at 3:19
    
Hi Qiaochu, its just that we haven't learned it yet. Can you explain to me how the orbit structure for specific choices of a and n works for this problem? THanks –  Buddy Holly Jan 24 '12 at 5:06
    
@Buddy: it would be a good idea to try to work this out for yourself. Try very small values of $a$ and $n$ first (e.g. try $a = 2$ and $n = 2, 3, 4$). –  Qiaochu Yuan Jan 24 '12 at 15:57

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