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Burnside Theorem

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Last updated date: 17th Apr 2024
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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

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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.

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FAQs on Burnside Theorem

1. What is a prime number?

A whole or a natural number higher than one that cannot be reduced exactly with any other whole number than itself and one (e.g., 2, 3, 5, 7, 11) is called a prime number. "Primality" is the quality of being a prime. Never forget that 1 is neither a prime number nor a composite number. Apart from one, the other numbers can all be categorised as prime and composite numbers while 2 is the smallest prime number and the sole even prime number.

2. What is an integer, and explain positive and negative integers?

Natural numbers, opposites, and zero are all examples of integers. An integer is a complete entity. Integers are numbers without a fractional component that can be positive, negative, or zero (no decimals). Integer examples can be 1, 2, 7, 8, 9, 12, etc. The letter "Z" stands for the totality of all integers.  Natural numbers are positive on the right-hand side of zero on the number line, while their opposites are negative numbers. The number line can be used to compare different integers.

3. What is representation theory?

Researchers analyse representations of algebras (groups, rings, and topological spaces) in representation theory by expressing their constituent parts as linear transformations of vector spaces. In more detail, a representation converts arbitrary algebraic concepts into matrices, making them more tangible. The significance of representation theory would be that it allows for the reduction of abstract problems to those that can be solved using the well-known mathematical theory of linear algebra. Furthermore, the style of vector space within which the algebra operates has a significant impact on representation theory.