An Overview
Burnside developed and proved Burnside's lemma in 1897, but historically speaking, Frobenius had already discovered it in 1887, and Cauchy had discovered this much prior, in 1845. As a result, it is also occasionally referred to as the "Cauchy-Frobenius Lemma". Based on internal symmetry, Burnside's lemma enables us to determine the number of equivalence classes in the sets.
In this article, we will learn more about the Burnside theorem and the Burnside theorem example.
History of William Burnside
William Burnside
Born: 2 July 1852
Died: 21 August 1927
Nationality: British
Contribution: He introduced the world to Burnside's theorem.
Statement of the Theorem
In group theory, Burnside's theorem asserts that group G is solvable if it is a finite group of order, where p and q are prime numbers, and a and b are non-negative integers. As a result, the order of every non-Abelian finite simple group is divisible by at least three unique primes.
Burnside Theorem Proof
$G$ is a simple group with a trivial centre, and $a$ is not zero.
Since $G$ is minimum, if it had a nontrivial proper normal subgroup\[H\], both \[H\] and \[\dfrac{G}{H}\] would be solvable, which would go against our presumption. So, $G$ is easy.
$G$ Would be a finite q-group, nilpotent, and solvable if they were equal to zero.
$G$ cannot also be abelian because it would then be solved. $G$ Centre must be basic because $G$ is simple.
There is an element \[g\] of \[G\] which has \[{{q}^{d}}\] conjugates, for some \[d>0\].
By the first statement of Sylow's theorem, G has a subgroup S of order pa because S is a nontrivial p-group, and its centre Z(S) is nontrivial. Fix a nontrivial element g\epsilon Z(S) . The number of conjugates of g is equal to the index of its stabiliser subgroup \[{{G}_{g}}\], which divides the index \[{{g}^{b}}\] of S (because S is a subgroup of \[{{G}_{g}}\]). Hence, this number is of the form \[{{q}^{d}}\]. Moreover, the integer d is strictly positive since g is nontrivial and therefore not central in G.
There exists a nontrivial irreducible representation \[\rho \] with character \[X\], such that its dimension \[n\] is not divisible by \[q\] and the complex number \[X(g)\] is not zero.
Let \[{{({{\chi }_{i}})}_{1~\le ~i~\le ~h}}\] be the family of irreducible characters of G over \[\sum\limits_{}^{} {_{{X_i}}} \], (here $({{X}_{1}})$ denotes the trivial character). Since $g$ is not in the same conjugacy class as 1, the orthogonality relation for the columns of the group's character table gives:
\[0 = \sum\limits_{i = 1}^h {{x_i}(1)} {x_i}(g) = 1 + \sum\limits_{i = 2}^h {{x_i}(1)} {x_i}(g)\]
Now the χi(g) are algebraic integers because they are sums of roots of unity. If all the nontrivial irreducible characters which don't vanish at g take a value divisible by q at 1, we deduce that
\[\dfrac{{ - 1}}{q} = \sum\limits_{i \ge 2,{x_i}(g) \ne 0}^{} {\dfrac{{{x_i}(1)}}{q}} {x_i}(g)\]
Burnside Lemma Applications
The number of unique ways that something can be done can be determined using Burnside's Counting Theory. This theory has intriguing implications in areas like switching theory and chemistry, secondary to its geometric applicability.
Burnside's Lemma and Polya's Enumeration Theorem can be used to determine the number of distinct colours present in an object.
Solved Examples
1. Determine how many different ways the four corners of a square may be coloured with two colours.
Ans: Naming each corner of the square, we have,
The number of ways to arrange the colours of the corner is ${{2}^{4}}=16$
${{D}_{4}}$ is the symmetries of the square.
${{R}_{0}}$ fixes all 16 arrangements.
${{R}_{90}}$ & ${{R}_{270}}$ fix arrangements with all 4 colours the same colour
Under ${{R}_{180}}\text{ }\!\!~\!\!\text{ : }\{1,3\}\text{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }\{2,4\}$
From these vertices $1$ and $3$ are of same colours $2$ and $4$ are of same colours
So, ${{R}_{180}}$ fixes $4$ colourings
Under $H$, it fixes $4$ colourings.
Under ${{n}^{3}}$ $D$, it fixes $8$ colourings (as ${{2}^{3}}=8$)
Similarly, D fixes $8$ colourings
$\dfrac{1}{\left| {{D}_{4}} \right|}\underset{\phi \in {{D}_{4}}}{\mathop \sum }\,|fix(\phi )|=\dfrac{1}{8}(16+2\cdot 2+4+2\cdot 4+2\cdot 8)=\dfrac{1}{8}(48)=6$
So, we have non-equivalent colours.
2. Determine how many different ways the four edges of a square may be coloured with six different colours.
Ans: The number of ways to arrange the colours of the edges of a square is ${{6}^{4}}$.
From Identity, we have ${{6}^{4}}$number of fixed colourings.
From the Rotation of order $4$, we have a $6$ number of fixed colourings.
From Rotation of order $2$, we have ${{6}^{2}}$ number of fixed colourings.
From edge Reflection, we have ${{6}^{3}}$ number of fixed colourings.
From Diagonal Reflection, we have ${{6}^{2}}$ number of fixed colourings.
From Burnside Theorem, we have,
$\dfrac{1}{\left| {{D}_{4}} \right|}\underset{\phi \in {{D}_{4}}}{\mathop \sum }\,|fix(\phi )|=\dfrac{1}{8}\left( {{6}^{4}}+2\cdot 6+{{6}^{2}}+2\cdot {{6}^{3}}+2\cdot {{6}^{2}} \right)=2310$
Therefore, $2310$ in total.
3. Let's say we cut a cake into six identical pieces. How many different ways can we decorate the cake with $n$ different colours if each piece gets one colour?
Ans: Total arrangement of colours is ${{n}^{6}}$.
The symmetry group of the cake is $G=\left\{ {{R}_{0}},{{R}_{60}},{{R}_{120}},{{R}_{180}},{{R}_{240}},{{R}_{300}} \right\}$
For rotations of order $6$ we have $\{1,2,3,4,5,6\}$($n$ colourings)
For rotations of order $3$ we have $\{1,3,5\}$&$\{2,4,6\}$( ${{n}^{2}}$colourings)
For \[{{R}_{180}}\] we have $\{1,4\},\{2,5\},\{3,6\}$( ${{n}^{3}}$colourings)
$\dfrac{1}{|G|}\underset{\phi \in G}{\mathop \sum }\,|fix(\phi )|=\dfrac{1}{6}\left( {{n}^{6}}+2n+2{{n}^{2}}+{{n}^{3}} \right).$
Important Formulae
\[\dfrac{{ - 1}}{q} = \sum\limits_{i \ge 2,{x_i}(g) \ne 0}^{} {\dfrac{{{x_i}(1)}}{q}} {x_i}(g)\]
- \[0 = \sum\limits_{i = 1}^h {{x_i}(1)} {x_i}(g) = 1 + \sum\limits_{i = 2}^h {{x_i}(1)} {x_i}(g)\]
Summary
Burnside's lemma is a result of group theory that can help when counting objects with symmetry taken into account. It gives a formula to count objects, where two objects that are related by a symmetry (rotation or reflection, for example) are not to be counted as distinct.
FAQs on Burnside Theorem
1. What is the statement of Burnside's Theorem, also known as the p^a q^b theorem?
In group theory, Burnside's Theorem states that if G is a finite group of order paqb, where p and q are prime numbers and a and b are non-negative integers, then the group G is solvable. A key implication of this is that the order of any non-Abelian finite simple group must be divisible by at least three distinct primes.
2. What is Burnside's Lemma and what is its main application?
Burnside's Lemma, also known as the Cauchy-Frobenius Lemma or the counting theorem, is a result in group theory used to count the number of distinct objects or configurations when symmetry is taken into account. Its primary application is in solving combinatorial problems by calculating the number of equivalence classes (or orbits) in a set under the action of a group. Essentially, it helps determine how many unique arrangements exist after accounting for rotations, reflections, and other symmetries.
3. Can you provide a simple example of using Burnside's Lemma?
A classic example is determining the number of unique ways to colour the four corners of a square using two colours (e.g., black and white). While there are 24 = 16 total possible colourings, many are identical under rotation or reflection. Burnside's Lemma is applied by considering the group of symmetries of the square (D4) and counting how many colourings are left unchanged (fixed) by each symmetry operation. The lemma provides a formula to average these counts, revealing that there are only 6 truly distinct colourings.
4. What is the key difference between Burnside's Theorem and Burnside's Lemma?
The primary difference lies in their purpose and scope within group theory:
- Burnside's Theorem (the paqb theorem) is a deep structural result about the properties of finite groups. It establishes a condition (having an order of paqb) that guarantees a group is solvable, which relates to its internal subgroup structure.
- Burnside's Lemma is a computational tool. It doesn't describe the structure of a group itself but uses the action of a group on a set to count distinct configurations or orbits. It is a cornerstone of combinatorial mathematics.
In short, the theorem is about a group's intrinsic properties, while the lemma is about a group's application in counting.
5. Why is the p^a q^b theorem so important in the study of finite simple groups?
Burnside's paqb theorem was a monumental step towards the classification of finite simple groups. By proving that any group with an order divisible by only two distinct primes is solvable, it automatically means such a group cannot be a non-Abelian simple group. This drastically narrows down the possible orders for these fundamental building blocks of group theory, forcing mathematicians to look for simple groups whose orders are divisible by at least three distinct primes (like the smallest non-Abelian simple group A5, of order 60 = 22 * 3 * 5).
6. How is Burnside's Lemma used in fields outside of pure mathematics, like chemistry?
Burnside's Lemma has significant practical applications, particularly in chemistry and computer science. In chemistry, it is used to count the number of isomers of a molecule. Isomers are compounds with the same chemical formula but different structures. By treating the atomic positions as a set and the molecule's rotational symmetries as the group, the lemma can calculate the number of structurally distinct molecules, avoiding overcounting due to spatial orientation.
7. What is a common mistake when applying Burnside's Lemma to a counting problem?
A common mistake is incorrectly identifying the set of elements fixed by a specific group action. For each symmetry operation (like a rotation or reflection), one must meticulously count the number of configurations that remain absolutely unchanged by that specific operation. Students often miscalculate this 'fixed set' size, especially for complex symmetries. Another frequent error is incorrectly defining the group of symmetries itself, either by missing a symmetry or including an invalid one.
