Discrete Mathematics lecture notes 13-1
December 6, 2013
33. Colorings and the cycle index
In the Lemma that is not Burnside’s, we now have a powerful tool for counting orbits of a permutation
group acting on a set. In these last two lectures, we’ll look at a specific type of set and specific type of
action: Instead of simply considering a permutation group G on a set X, we will examine how G acts on
the set of colorings of X, and as a result obtain a formula for the number of essentially distinct colorings
possible. The advantage of this point of view is that we will unify what had formerly been several seemingly
unrelated computations, hinting at a deeper structure encoded within the group theory we’ve developed.
Let’s start out by fixing a set X of size n and a permutation group G ≤ ΣX . We want to color the points
of X, say red, blue, green, . . . , and ask how many distinct colorings are possible, up to the symmetries of G.
For example, if X is the corners of the square and we have only the colors black and white to choose from,
there are a total of 24 = 16 different possible colorings:
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Here we’ve been careful in our organization of the colorings: Each row consists of the same number of black
and white dots, but more importantly, they are arranged in such a way that one one can pass from one
coloring to another via some element of D8 , the symmetry group of the square.1 Let’s generalize this.
1 Indeed,
within each row, passing to the right corresponds to the 90◦ clockwise rotational symmetry.
1
In addition to fixing X of size n and a permutation group G, let’s also fix a set of r “colors,” called
f1 , f2 , . . . , fr .2 Let F denote this set of r colors.
Definition 1. An r-coloring of X is a function ω : X → F . We denote by Ω the set of all r-colorings of X.
Exercise. Show that |Ω| = rn .
Our goal is to now understand how G can transform one coloring into another, and then count the number
of G-orbits of this action. We will do this by first explicitly turning G into a group of symmetries of the set
Ω.
Definition 2. For g ∈ G, let gb ∈ ΣΩ be the permutation of Ω defined by
(b
g (ω))(x) := ω(g −1 (x)).
b ⊆ ΣΩ denote the set of all permutations of Ω that arise in this manner: G
b = {b
Let G
g | g ∈ G}.
This definition is a bit convoluted; the key to understanding it is remembering where each of the pieces
live. x ∈ X is a point of the underlying set. ω is a coloring of X, so ω(x) ∈ F is a color. g ∈ Σx is a
permutation of x, so g −1 is as well, hence g −1 (x) ∈ X is a point and ω(g −1 (x)) ∈ F is a color. Finally, gb,
what we’re defining, is a permutation of Ω: When fed a coloring ω, it returns another coloring gb(ω) (which
is in turn only understood by describing where points of X are sent).
One might reasonably ask why gb is defined using g −1 instead of g. An example-based explanation would
be to note that we’ve already seen this in the colorings of the square used above: If the vertices of the square
are labeled
x2
x1
x4
x3
◦
then the clockwise 90 rotation R is given by the cycle (x1 x2 x3 x4 ). If we examine the second row above,
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ω2
ω1
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ω4
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the idea is that ω2 is obtained from ω1 by applying R to the underlying square, which we express by saying
b 1 ). Similarly ω3 = R(ω
b 2 ), ω4 = R(ω
b 3 ), and ω1 = R(ω
b 4 ). What does this actually mean in terms of
ω2 = R(ω
colorings, though? If we use the symbol b for the color black and w for the color white, we can express all
the colorings in the table below:
x1 x2 x3 x4
ω1 w b
b
b
ω2 b
w b
b
ω3 b
b
w b
ω4 b
b
b
w
b 1 ) send xi , in terms of the rotation R = (x1 x2 x3 x4 ) and ω1 ? Since ω2 (x2 ) = w
Now, where does ω2 = R(ω
and the only point colored white by ω1 is x1 , we see that ω2 colors by first undoing the action of R and the
coloring via ω1 . As R(x1 ) = x2 , we have x1 = R−1 (x2 ), so
w = ω1 (x1 ) = ω1 (R−1 (x2 )) = ω2 (x2 ),
b 1 ),
which we in turn generalize to say ω2 (xi ) = ω1 (R−1 (xi )). From the original observation that ω2 = R(ω
we see that we’ve actually recreated the strange looking definition of gb given above.
There is a deeper reason for defining gb(ω) in terms of g −1 , which is the content of the following:
2 By definition, a color is just one of these f . We use the letter f instead of the more suggestive (unless you’re German,
i
say) c for color because we will be talking about cycles shortly.
2
Proposition 3. For all g, h ∈ G:
(a) gb ◦ b
h = gd
◦ h.
(b) (b
g)
−1
−1 .
= gd
Proof. It’s important to remember where the compositions and inversions of this Proposition are taking
place: For part (a), in the expression gb ◦ b
h the composition is in ΣΩ , while in g[
◦ h it is in ΣX .3 The content
b : g 7→ gb is a group homomorphism, i.e., a set-map
of this result is summarized with the phrase b : G → G
that respects the multiplicative structure of groups.
(a) We must show (b
g◦b
h)(ω) = g[
◦ h(ω) for all ω ∈ Ω, which in turn means we need
?
((b
g◦b
h)(ω))(x) = (g[
◦ h(ω))(x)
for all ω ∈ Ω and x ∈ X. Using the definition of b, the left hand side becomes
((b
g◦b
h)(ω))(x) = (b
g (b
h(ω)))(x) = (b
h(ω))(g −1 (x)) = ω(h−1 (g −1 (x))) = ω((h−1 ◦ g −1 )(x)),
while the right is
(g[
◦ h(ω))(x) = ω((g ◦ h)−1 (x)).
Since we know for all g, h ∈ G that (g ◦ h)−1 = h−1 ◦ g −1 , the result follows.
−1 = id .4 By part (a), we know
(b) By definition of the inverse, this amounts to showing gb ◦ gd
Ω
−1 = g\
d
gb ◦ gd
◦ g −1 = id
X.
d
Show from the definition of b that id
X = idΩ .
Exercise. If we had defined gb(ω) using g instead of g −1 , show that part (a) of the previous Proposition would
not hold.5
b is a surjection. In fact, it is an injection as well:
By definition, the map G → G
Proposition 4. If r ≥ 2 and g, h ∈ G are such that gb = b
h, then g = h.6
Proof. gb = b
h iff for all ω ∈ Ω we have gb(ω) = b
h(ω) iff for all ω ∈ Ω and x ∈ X we have
(b
g (ω))(x) = (b
h(ω))(x) ⇔ ω(g −1 (x)) = ω(h−1 (x)).
I claim that this means g −1 (x) = h−1 (x) for all x ∈ X: If not, then for some x0 such that g −1 (x0 ) 6= h−1 (x0 ),
there is a coloring ω0 with ω0 (g −1 (x0 )) = f1 and ω0 (h−1 (x0 )) = f2 , contrary to our assumption that gb = b
h.
But if g −1 (x) = h−1 (x) for all x ∈ X, we have g −1 = h−1 ∈ ΣX , and hence g = h, as desired.
b of Ω, but G
In other words, not only does the permutation group G of X induce a permutation group G
b are essentially the same group.7
and G
3 Since
g ∈ ΣX and g
b ∈ ΣΩ , this is in fact the only interpretation that makes sense.
−1 ◦ g
we also need that gd
b = idΩ as well; can you show in general that this second equality follows from the first?
5 This is actually a reflection of the “left action vs. right action” dichotomy hinted at in the previous notes.
6 What happens in the degenerate case r = 1? Recall that r is the number of colors in our colorings.
7 We say that the bijective homomorphism G → G
b is an isomorphism of groups. The existence of an isomorphism means
b are two different representations of the same abstract group, which should be thought of as a platonic ideal of a
that G and G
set of symmetries, divorced from the mundane reality of the object being symmatrized.
4 Actually,
3
Ok, let’s step back. The goal originally was to count the number of colorings of X, up to the symmetries
induce by the permutation group G. We’ve now described explicitly how G acts on the set of colorings Ω,
b on Ω: We seek a description
and in so doing rephrased the question as calculating the number of orbits of G
b
of |G\Ω|. Of course, we already have such a numerical description in the Lemma that is not Burnside’s:
b
|G\Ω|
=
1 X gb
1 X gb
|Ω | =
|Ω |
b
|G|
|G|
g∈G
b
g
b∈G
b is a bijection, so it
Here, the second expression is just a reformulation of the result that the map G → G
b The question is now whether we can improve on
does not matter whether we choose to index over G or G.
this expression by, for instance, finding a way to deal with r- and r0 -coloring simultaneously.
Let’s consider, for fixed g ∈ G, the set Ωgb. If ω is a fixed point of gb, then by definition gb(ω) = ω, or for
all x ∈ X
ω(g −1 (x)) = ω(x).
If we express g in terms of its disjoint cycles,
g = (x11 x12 . . . x1`1 )(x21 x22 . . . x2`2 ) . . . (xk1 xk2 . . . xk`k ),
then an easy induction shows that
ω(xi1 ) = ω(xi2 ) = . . . = ω(xi`i )
for all 1 ≤ i ≤ k. In other words, ω ∈ Ωgb if and only if ω is constant on each of g’s cycles. As we can choose
any of the r colors of F for each cycle of g, we have proved
Lemma 5. If g ∈ G has k cycles, then |Ωgb| = rk .
We will see in the next lecture that it is worthwhile to distinguish between the lengths of the various
cycles, so that we are lead to define:
Definition 6. For g ∈ G, the cycle type of g is the expression [1c1 2c2 . . . ncn ], where ci is the number of
cycles of length i of g.
Note that for all g ∈ G, we have
n
P
i · ci = n. If g has k cycles, then
i=1
n
P
ci = k.
i=1
We wish to manipulate cycle types algebraically, so we introduce the formal variables x1 , x2 , . . . , xn , and
define ζg (x1 , . . . , xn ) to be the polynomial xc11 xc22 . . . xcnn .
As the xi are formal variables (being manipulated like variables in a polynomial ring, because that’s what
they are) we can evaluate ζg (x1 , . . . , xn ) at an n-tuple of real numbers (a1 , . . . , an ) ∈ Rn by plugging in ai
for xi . If we evaluate at the constant n-tuple (r, r, . . . , r) (r is the number of colors in F ), we have
ζg (r, r, . . . , r) = rc1 rc2 . . . rcn = rc1 +c2 +...+cn = rk = |Ωgb|.
Definition 7. The cycle index of G is the polynomial
ζG (x1 , . . . , xn ) :=
1 X
ζg (x1 , . . . , xn ).
|G|
g∈G
If we evaluate ζG at (r, r, . . . , r), we get
ζG (r, r, . . . , r) =
1 X gb
1 X
b
ζg (r, r, . . . , r) =
|Ω | = |G\Ω|,
|G|
|G|
g∈G
g∈G
and we have proved
Theorem 8. The number of G-distinct r-colorings of X is ζG (r, r, . . . , r).
4
Example 9. Let’s look at the square
1
2
4
3
with group of symmetries D8 generated by R = (1234) and F = (24). In terms of this labeling, the elements
of D8 , with their corresponding cycle indices are given by:8
g
id
R
R2
R3
F
FR
F R2
F R3
cycles
(1)(2)(3)(4)
(1234)
(13)(24)
(1432)
(1)(24)(3)
(14)(23)
(13)(2)(4)
(12)(34)
ζg
x41
x14
x22
x14
x21 x12
x22
x21 x12
x22
and we compute
1 4
(x + 3x22 + 2x4 + 2x21 x2 ).
8 1
Evaluating each xi at 1, we conclude that there is
ζD8 =
ζD8 (1, 1, 1, 1) =
1
(1 + 3 + 2 + 2) = 8
8
distinct ways to color the edges of a square with a single color.9 If we we want to ask how many distinct
2-colorings are possible, we find:
ζD8 (2, 2, 2, 2) =
1
(16 + 3 · 4 + 2 · 2 + 2 · 8) = 6,
8
which agrees with our opening example (count the number of rows of colored squares on the first page).
What’s nice is that we get for free the number of 3-colorings, and 13-colorings, and etc. in the same manner:
ζD8 (3, 3, 3, 3)
=
ζD8 (13, 13, 13, 13)
=
1
(81 + 3 · 9 + 2 · 3 + 2 · 27) = 21
8
1
(134 + 3 · 132 + 2 · 13 + 2 · 133 ) = 3679
8
..
.
8 While we normally suppress 1-cycles from our notation, in this context it is important not to forget that they have never
left us.
9 This should not be surprising. Actually, we haven’t explicitly proved that this formula works for the case r = 1, as
Proposition 4 assumed that r ≥ 2, and this was actually a central technical piece of the argument. Can you give a separate
argument to show that ζG (1, 1, . . . , 1) = 1 in general?
5