Burnside's Lemma • Sons Only Take After Their Fathers' Negative Attributes

Why Another Burnside’s Lemma Article?

Most explanations of Burnside’s Lemma start with “Let $G$ be a finite group acting on a finite set $X$ …” and before you know it, you’re drowning in orbits, stabilizers, and group actions. That’s great if you’re studying abstract algebra, but what if you’re just trying to count necklaces?

Here’s what Burnside’s Lemma actually is: intentional double counting. That’s it. No group theory required.

The Core Idea

Imagine you have a collection of objects with symmetries (like necklaces that can be rotated, or dice that look the same after certain rotations). You want to count how many “truly different” objects there are.

The way you’d really love to do this is to:

Count the total number of objects ignoring any symmetries. Usually this is pretty simple since you can use basic ideas like the Fundamental Principle of Counting.
Count the number of symmetries there are. Also usually pretty simple.
Divide the two.

This makes sense because in Step 1 you’re obviously double counting many cases because when you take a particular case and rotate it, reflect it, etc., you’ll typically end up with one or more other cases that you’ve already counted. To account for this, you’d like to divide by how many ways there are to rotate and reflect and whatnot to fix the double counting.

There’s just one problem: not every case is double counted the same number of times. Some special cases might be double counted once for every possible symmetry. Others might not be double counted at all!

How do we fix this? Enter Burnside’s Lemma. The trick to Burnside’s Lemma is that instead of counting like in Step 1 where every case is double counted a different number of times, we count in such a way that every case is double counted exactly the same number of times. At first glance this seems like we’re going in the wrong direction - in Step 1 we already have some double counting, why would we choose to count so there’s even more double counting? The idea is that if we know we’ve double counted every case by 8x (if there are 8 symmetries), then dividing by 8 will give you the number of “truly different” cases.

In practice there are two ways we could intentionally double count everything the same number of times:

Method 1: Look at each distinct object and ask “how many times am I counting this?” (This is usually the number of symmetries that leave it unchanged.)

Method 2: Look at each symmetry and ask “how many objects stay the same under this symmetry?”

Both methods count the same total, but Method 2 is usually the simplest. This is usually what people are referring to when they say Burnside’s Lemma.

Warm-up: Bracelets

Let’s make bracelets with 3 beads, each either red ● or blue ●. A bracelet can be rotated or flipped over, so these patterns can look the same after transformation.

Without Burnside

We could list all $2^3 = 8$ possible bead arrangements. Let’s visualize them as triangular bracelets:

I’ve grouped together the bracelets that are the same - those that can be rotated or flipped to match each other. Clearly there are only 4 distinct bracelets: , , , and .

Burnside Method 1: Count by Object

A triangular bracelet has 6 symmetries: identity, two rotations (120° and 240°), and three reflections through each vertex. For each of the 8 bracelet patterns, count how many of these symmetries leave it unchanged:

: Unchanged by all 6 symmetries → 6
: Unchanged by identity + one reflection → 2
: Unchanged by identity + one reflection → 2
: Unchanged by identity + one reflection → 2
: Unchanged by identity + one reflection → 2
: Unchanged by identity + one reflection → 2
: Unchanged by identity + one reflection → 2
: Unchanged by all 6 symmetries → 6

Total: $6 + 2 + 2 + 2 + 2 + 2 + 2 + 6 = 24$ ✓

Burnside’s Formula: $\text{Distinct bracelets} = \frac{24}{6} = 4$

As described above, all we really did here was find a systematic way to make sure we double counted every “truly different” bracelet 6 times. Here’s a visual breakdown, where I’ve grouped everything nicely for convenience:

Burnside Method 2: Count by Symmetry

Here’s where things get powerful. Figuring out how much to double count each object works, but requires examining every object. The downside is that even simple problems can have many, many objects to check.

A cleverer approach is to look at each symmetry and count how many objects we would double count under that symmetry. Since there are usually far fewer symmetries than objects, and we can use basic counting principles for each symmetry, this method is usually more efficient. Let’s do that below:

Symmetry	Count	Patterns that stay unchanged
Identity	8	All 8 patterns:
Rotate 120°	2	Only all-same-color patterns:
Rotate 240°	2	Same as 120° rotation:
Reflect through top vertex	4	Bottom-left and bottom-right beads must match:
Reflect through bottom-left vertex	4	Top and bottom-right beads must match:
Reflect through bottom-right vertex	4	Top and bottom-left beads must match:

Notice that we’re counting the same number of each thing as in Method 1, but we’ve arrived there by a different road.

Burnside’s Formula: $\text{Distinct bracelets} = \frac{1}{6}(8 + 2 + 2 + 4 + 4 + 4) = \frac{24}{6} = 4$

A Bigger Necklace

Let’s make a 5-bead necklace with 3 colors: red ●, blue ●, and green ●. The necklace can be rotated but not flipped.

Without Burnside

With 5 beads and 3 colors, there are $3^5 = 243$ total arrangements. Trying to list them all and group equivalent ones would be a nightmare!

Burnside Method 2: Count by Symmetry

A 5-bead necklace (rotation only) has 5 symmetries: identity and rotations by 72°, 144°, 216°, and 288°.

Symmetry	Count	Description
Identity	243	All $3^5 = 243$ arrangements stay unchanged
Rotate 72° (one bead)	3	Pattern must repeat every 5 positions. Since 5 is prime, only monochromatic works:
Rotate 144° (two beads)	3	Same reasoning as 72° rotation
Rotate 216° (three beads)	3	Same reasoning as 72° rotation
Rotate 288° (four beads)	3	Same reasoning as 72° rotation

Burnside’s Formula: $\text{Distinct necklaces} = \frac{1}{5}(243 + 3 + 3 + 3 + 3) = \frac{255}{5} = 51$

Notice how much easier this was than trying to list all 243 arrangements and group them! The key insight is that 5 is prime, so non-identity rotations only fix monochromatic necklaces.

Another way of thinking about this is that among the 243 objects, every case is double counted exactly 5 times except the 3 cases where all beads are the same color. So before we divide by 5 we need to first add $3\times4=12$ to 243 so that EVERYTHING is double counted 5 times.

2017 AMC 10B Problem 18

In the figure below, 3 of the 6 disks are to be painted blue, 2 are to be painted red, and 1 is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?

(A) 6 (B) 8 (C) 9 (D) 12 (E) 15

Solution

Symmetry	Count	Description
Identity	60	All $\frac{6!}{3!2!1!} = 60$ colorings stay unchanged
120° rotation	0	Creates two 3-cycles (corners and edges). Need monochromatic cycles, but can’t split 3B, 2R, 1G into two groups of 3
240° rotation	0	Same reasoning as 120° rotation
Vertical reflection	4	Middle disks must match (2B or 2R), bottom corners must match (the other pair), top and bottom-middle get remaining colors. 2 choices for pairs × 2 for singles
Other reflection 1	4	By symmetry with vertical reflection
Other reflection 2	4	By symmetry with vertical reflection

Burnside’s Formula: $\text{Distinct paintings} = \frac{1}{6}(60 + 0 + 0 + 4 + 4 + 4) = \frac{72}{6} = 12$

The answer is $\boxed{\textbf{(D) } 12}$ .

2021 AMC 10B Fall Problem 24

A cube is constructed from 4 white unit cubes and 4 blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.) (A) 7 (B) 8 (C) 9 (D) 10 (E) 11

Solution

What’s a little less obvious here is that there are 24 rotational symmetries. One way to see this is to consider a particular vertex. Including the identity transformation, there are 8 vertices it could move to. From there you could “twist” the cube about that new vertex in three different ways. Hence 24.

Symmetry	Count	Description
Identity	70	All $\binom{8}{4} = 70$ arrangements
Face rotations (90°, 270°)	12	Creates two 4-cycles. Each must be monochromatic. With 4W and 4B, one cycle gets white, other blue. 2 colorings × 6 rotations
Face rotations (180°)	18	Creates 4 pairs of opposite corners. Each pair must match. $\binom{4}{2} = 6$ ways to choose 2 pairs for white × 3 rotations
Vertex rotations (120°, 240°)	32	Creates two 3-cycles and fixes 2 corners. One 3-cycle gets 3W, other gets 3B, fixed corners get 1W and 1B. 2 ways to assign cycles × 2 ways for corners × 8 rotations
Edge rotations (180°)	36	Creates 4 pairs. $\binom{4}{2} = 6$ ways to choose which pairs get white × 6 rotations

Burnside’s Formula: $\text{Distinct cubes} = \frac{1}{24}(70 + 12 + 18 + 32 + 36) = \frac{168}{24} = 7$

The answer is $\boxed{\textbf{(A) } 7}$ .

2007 AMC 12B Problem 16

Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?

(A) 15 (B) 18 (C) 27 (D) 54 (E) 81

Solution

Like the cube in the previous problem, we can see that there are 12 rotations to consider because a vertex could be “sent” to any of the 4 vertices, and then twisted around that vertex in one of 3 ways.

Symmetry	Count	Description
Identity	81	All $3^4 = 81$ arrangements
Face rotations (120°, 240°)	72	Three faces in cycle must be monochromatic, fixed face can be any color. $3 \times 3 = 9$ colorings × 8 rotations
Edge rotations (180°)	27	Two pairs of adjacent faces must each be monochromatic. $3 \times 3 = 9$ colorings × 3 rotations

Burnside’s Formula: $\text{Distinct tetrahedra} = \frac{1}{12}(81 + 72 + 27) = \frac{180}{12} = 15$

The answer is $\boxed{\textbf{(A) } 15}$ .

1990 AJHSME (AMC 8) Problem 25

How many different patterns can be made by shading exactly two of the nine squares? Patterns that can be matched by flips and/or turns are not considered different. For example, the patterns shown below are not considered different.

(A) 3 (B) 6 (C) 8 (D) 12 (E) 18

Solution

Symmetry	Count	Description
Identity	36	All $\binom{9}{2} = 36$ possible patterns
90° rotation	0	Impossible
270° rotation	0	Impossible
180° rotation	4	Two shaded squares must form opposite pairs (2 corner pairs + 2 edge pairs)
Vertical reflection	6	$\binom{3}{2}=3$ ways to choose 2 squares on the middle axis + $\binom{3}{2}$ ways to choose 2 squares surrounding the axis
Horizontal reflection	6	Same as the vertical reflection
Diagonal reflection (main)	6	Same as the vertical reflection
Diagonal reflection (anti)	6	Same as the vertical reflection

Burnside’s Formula: $\text{Distinct patterns} = \frac{1}{8}(36 + 0 + 0 + 4 + 6 + 6 + 6 + 6) = \frac{64}{8} = 8$

The answer is $\boxed{\textbf{(C) } 8}$ .

2000 AMC 12 Problem 25

Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)

(A) 210 (B) 560 (C) 840 (D) 1260 (E) 1680

Solution

To figure out how many rotational symmetries there are, we use the same idea as we’ve been using: a vertex can be mapped to any of the 6 vertices and can then be “twisted” 4 different ways. So 24 rotations.

Symmetry	Count	Description
Identity	40320	All $8! = 40320$ arrangements
All 23 non-identity rotations	0	Since every face is a different color, no non-identity rotation leaves any arrangement unchanged

Burnside’s Formula: $\text{Distinct octahedra} = \frac{1}{24}(40320 + 0) = 1680$

The answer is $\boxed{\textbf{(E) } 1680}$ .