Group Actions Notes - Burnside's Theorem
Burnside's Theorem
Burnside's Theorem holds significant importance in the study of Abstract Algebra and Group Theory. Introduced in the late 19th century, this theorem simplifies the counting of distinct objects under symmetry operations. Its applications span algebraic structures, combinatorics, and computational mathematics, making it a pivotal topic for Mathematics.
What You Will Learn?
In this post, you will explore:
- Definition: Fixed by an Element
- Definition: Fixed by a Group
- Burnside’s Theorem: Let \((G,\circ )\) be a finite group and \(S\) be a non-empty finite set. If \(S\) is a G-Set then the number of orbits of \(G\) is \begin{align*}\frac{1}{\big|G\big|}\displaystyle\sum_{g\in G}^{}F(g)\end{align*}, where \(F(g)\) is the number of elements of \(S\) fixed by \(g\).
- Theorem-2: Let \((G,\circ )\) be a finite group and \(S\) be a non-empty finite set and \(S\) be a G-Set. Let \(S_{0}=\set{a\in S: g\cdot a=a~\forall~\in G}\). If \(p^{n}\) divides \(\big|G\big| \) where \(p\) is a prime, then \begin{align*} |S|\equiv |S_{0}|\big(mod~p \big) \end{align*} Also in addition if \(p \) does not divide \(|S|\) then \(S_{0}\ne \phi\).
- Theorem-3: Let \((G,\circ )\) be a finite group and \(H\) be a subgroup of \(G\) such that \(|H|=p^{k}\) where \(p\) is a prime and \(k\in \mathbb{Z}^{+}\).
- \( \big[G:H \big] \equiv \big[ N(H) :H \big] \big(mod~p \big)\).
- If \(p\big| \big[G:H \big] \) then \(N(H)\ne H \).
Things to Remember
- Set Theory
- Relations
- Mappings
- Group Theory
Introduction
Burnside’s Theorem is a cornerstone in Mathematics, especially in the realm of group actions. It provides a powerful tool for enumerating distinct configurations by leveraging symmetry. This theorem is widely taught across various universities due to its practical applications in solving complex problems in abstract algebra and combinatorics.
Fixed by an Element
Definition:
Let \((G,\circ )\) be a group and \(S\) be a non-empty set and \(S\) be a G-Set. Let \(a\in S\) and \(g\in G\). Then \(a\) is said to be fixed by \(g\) if \(g\cdot a =a\).
Fixed by a Group
Definition:
Let \((G,\circ )\) be a group and \(S\) be a non-empty set and \(S\) be a G-Set. Let \(a\in S\). Then \(a\) is said to be fixed by \(G\) if \(g\cdot a =a,~\forall g\in G\).
Burnside's Theorem
Statement:
Let \((G,\circ )\) be a finite group and \(S\) be a non-empty finite set. If \(S\) is a G-Set then the number of orbits of \(G\) is \begin{align*}\frac{1}{\big|G\big|}\displaystyle\sum_{g\in G}^{}F(g)\end{align*}, where \(F(g)\) is the number of elements of \(S\) fixed by \(g\).
Proof:
Given that \((G,\circ )\) is a finite group and \(S\) is a non-empty finite set.
Let \(S\) is a G-Set.
Then there exists an action \(\cdot :G\times S \to S\) of \(G\) on \(S\).
Let \(F(g)\) is the number of elements of \(S\) fixed by \(g\in G\). Then we have for \(g\in G\)
\begin{align*}
S_{g}=\set{a\in S: g\cdot a=a}
\end{align*}
such that \(|S_{g}|=F(g)\).
Also we have for \(a\in S\)
\begin{align*}
G_{a}=\set{g\in G: g\cdot a=a}
\end{align*}
Let
\begin{align*}
T=\set{(g,a)\in G\times S: g\cdot a=a}
\end{align*}
Then we have
\begin{align}
& |T|=\displaystyle\sum_{g\in G}^{}|S_{g}| \text{ and } |T|=\displaystyle\sum_{a\in S}^{}|G_{a}|\nonumber\\
\implies & |T|=\displaystyle\sum_{g\in G}^{}F(g) \text{ and } |T|=\displaystyle\sum_{a\in S}^{}|G_{a}| \nonumber\\
\implies & \displaystyle\sum_{g\in G}^{}F(g) = \displaystyle\sum_{a\in S}^{}|G_{a}|
\end{align}
Let \(k\) be the number of orbits of \(G\).
Let \(S=[a_{1}]\cup [a_{2}]\cup … \cup [a_{k}]\) where
\([a_{1}], [a_{2}], … , [a_{k}]\) are all distinct equivalence classes of \(S\).
Let \(x,y\in [a_{p}]\) then we have \([x]=[y]=[a_{p}]\) and then \(\big|[x]\big|=\big|[y]\big|=\big|[a_{p}]\big|\).
Again
\begin{align}
&\big[G:G_{a} \big]=\big|[a] \big| \nonumber\\
\implies & \frac{\big|G \big|}{\big|G_{a} \big|}=\big|[a] \big| \nonumber\\
\implies & \big| G_{a}\big|=\frac{\big|G \big|}{\big|[a] \big|} \nonumber\\
\implies & \displaystyle\sum_{a\in [a_{p}]}^{}\big| G_{a}\big|=\displaystyle\sum_{a\in [a_{p}]}^{}\frac{\big|G \big|}{\big|[a] \big|} \nonumber\\
\implies & \displaystyle\sum_{a\in [a_{p}]}^{}\big| G_{a}\big|=\displaystyle\sum_{a\in [a_{p}]}^{}\frac{\big|G \big|}{\big|[a_{p}] \big|} \nonumber\\
\implies & \displaystyle\sum_{a\in [a_{p}]}^{}\big| G_{a}\big|=\frac{\big|G \big|}{\big|[a_{p}] \big|}\displaystyle\sum_{a\in [a_{p}]}^{}1 \nonumber\\
\implies & \displaystyle\sum_{a\in [a_{p}]}^{}\big| G_{a}\big|=\frac{\big|G \big|}{\big|[a_{p}] \big|}\big|[a_{p}] \big| \nonumber\\
\implies & \displaystyle\sum_{a\in [a_{p}]}^{}\big| G_{a}\big|=\big|G \big|
\end{align}
Now from (1),
\begin{align*}
&\displaystyle\sum_{g\in G}^{}F(g) = \displaystyle\sum_{a\in S}^{}|G_{a}| \\
\implies &\displaystyle\sum_{g\in G}^{}F(g) = \displaystyle\sum_{a\in [a_{1}]}^{}|G_{a}|+\displaystyle\sum_{a\in [a_{2}]}^{}|G_{a}|+…+\displaystyle\sum_{a\in [a_{k}]}^{}|G_{a}| \\
\implies &\displaystyle\sum_{g\in G}^{}F(g) = \big|G \big|+\big|G \big|+…+\big|G \big| \text{ using (2)}\\
\implies &\displaystyle\sum_{g\in G}^{}F(g) = k\big|G \big|\\
\implies & k=\frac{1}{\big|G \big|}\displaystyle\sum_{g\in G}^{}F(g) \\
\end{align*}
Theorem-2
Statement:
Let \((G,\circ )\) be a finite group and \(S\) be a non-empty finite set and \(S\) be a G-Set. Let \(S_{0}=\set{a\in S: g\cdot a=a~\forall~g\in G}\). If \(p^{n}\) divides \(\big|G\big| \) where \(p\) is a prime, then \begin{align*} |S|\equiv |S_{0}|\big(mod~p \big) \end{align*} Also in addition if \(p \) does not divide \(|S|\) then \(S_{0}\ne \phi\).
Proof:
Given that \((G,\circ )\) is a finite group and \(S\) is a non-empty finite set and \(S\) is a G-Set.
Then there exists an action \(\cdot :G\times S \to S\) of \(G\) on \(S\).
Also we have \(S_{0}=\set{a\in S: g\cdot a=a~\forall~g\in G}\).
Let \(p^{n}\) divides \(\big|G\big| \) where \(p\) is a prime.
To prove \(|S|\equiv |S_{0}|\big(mod~p \big) \)
Let
\begin{align*}
& x\in S_{0} \\
\implies & g\cdot x=x~\forall~g\in G \\
\implies & G_{x}=G \\
\implies & \big[G: G_{x} \big]=1 \\
\end{align*}
And let
\begin{align*}
& y\in A-S_{0} \\
\implies & g\cdot x=x~\text{ for some } g\in G \\
\implies & G_{x}\ne G \\
\implies & \big[G:|G_{x}| \big]\gt 1 \\
\implies & \frac{|G|}{|G_{x}|} \gt 1 \\
\implies & |G| \gt |G_{x}| \\
\end{align*}
But \(G_{x}\) is a subgroup of \(G\) then \(p\) divides \(\big|G_{x} \big|\). Therefore \(p\) divides \( \frac{|G|}{|G_{x}|} =\big[G: |G_{x}| \big] ~\forall~x\in A-S_{0}\) implies \(p\) divides \( \displaystyle\sum_{a\in A- S_{0}}^{}\big[G:G_{a} \big]\).
We have
\begin{align*}
|S|= \displaystyle\sum_{a\in A}^{}\big[G:G_{a} \big]
\end{align*}
where \(A\) is a subset of \(S\) containing exactly one member from each orbit \([a]\) of \(G\).
Now
\begin{align*}
& |S|= \displaystyle\sum_{a\in S_{0}}^{}\big[G:G_{a} \big]+ \displaystyle\sum_{a\in A- S_{0}}^{}\big[G:G_{a} \big]\\
\implies & |S|= \displaystyle\sum_{a\in S_{0}}^{} 1 + \displaystyle\sum_{a\in A- S_{0}}^{}\big[G:G_{a} \big]\\
\implies & |S|= |S_{0}| + \displaystyle\sum_{a\in A- S_{0}}^{}\big[G:G_{a} \big]\\
\implies & |S|-|S_{0}|=\displaystyle\sum_{a\in A- S_{0}}^{}\big[G:G_{a} \big]\\
\end{align*}
Therefore \(p\) divides \(|S|-|S_{0}| \).
Hence \(|S|\equiv |S_{0}|\big(mod~p \big) \).
2nd Part-
Let \(p \) does not divide \(|S|\) then we have
\begin{align*}
& |S|\equiv |S_{0}|\big(mod~p \big)\\
\implies & p \big| |S|-|S_{0}| \\
\implies & p \bcancel{|} |S_{0}| \\
\implies & |S_{0}|\ne 0 \\
\implies & S_{0}\ne \phi \\
\end{align*}
This proves the theorem.
Theorem-3
Statement:
Let \((G,\circ )\) be a finite group and \(H\) be a subgroup of \(G\) such that \(|H|=p^{k}\) where \(p\) is a prime and \(k\in \mathbb{Z}^{+}\).
- \( \big[G:H \big] \equiv \big[ N(H) :H \big] \big(mod~p \big)\).
- If \(p\big| \big[G:H \big] \) then \(N(H)\ne H \).
Proof:
Given that \((G,\circ )\) is a finite group and \(H\) is subgroup of \(G\) such that \(|H|=p^{k}\) where \(p\) is a prime and \(k\in \mathbb{Z}^{+}\).
To prove \( \big[G:H \big] \equiv \big[ N(H) :H \big] \big(mod~p \big)\)
Let \(S=\set{aH: a\in G}\).
Then
\begin{align}
|S|=\big[G:H\big]
\end{align}
Let us define an action \(\cdot :H\times S \to S\) of \(H\) on \(S\) such that \(h\cdot aH=\big(h\circ a \big)H \) for \( h\in H\) and \(a\in G \).
Let \(S_{0}=\set{aH\in S: h\cdot aH=aH~\forall~h\in G}\). Then
\begin{align}
|S|\equiv |S_{0}|\big(mod~p \big)
\end{align}
Let \(T=\set{yH: y\in N(H)}\). Then
\begin{align}
|T|=\big[N(H):H\big]
\end{align}
- To prove \(S_{0}\subseteq T\)
let \begin{align*} & xH \in S_{0} \text{ for some } x \in G \\ \implies & h\cdot xH= xH~\forall~h\in H \\ \implies & \big(h\circ x \big)H= xH~\forall~h\in H \\ \implies & \big[x^{-1}\circ\big(h\circ x \big)\big]H= H~\forall~h\in H \\ \implies & x^{-1}\circ h\circ x \in H~\forall~h\in H \\ \implies & x^{-1}\circ H\circ x \subseteq H \\ \implies & x^{-1}\circ H\circ x = H ~\big[ \because H\subseteq x^{-1}\circ H\circ x \big]\\ \implies & x \in N(H) \\ \implies & xH \in T \\ \end{align*} \(\therefore S_{0}\subseteq T \). - To prove \(T \subseteq S_{0}\)
let \begin{align*} & yH \in T \text{ for some } y \in N(H) \\ \implies & y^{-1}\circ H\circ y = H \\ \implies & y^{-1}\circ H\circ y \subseteq H \\ \implies & y^{-1}\circ h\circ y \in H ~\forall~h\in H \\ \implies & \big(y^{-1}\circ h\circ y \big)H=H ~\forall~h\in H \\ \implies & \big(h\circ y \big)H=yH ~\forall~h\in H \\ \implies & h\cdot yH=yH ~\forall~h\in H \\ \implies & yH\in S_{0} \end{align*} \(\therefore T \subseteq S_{0}\).
Therefore
\begin{align}
& T = S_{0} \nonumber\\
\implies & |S_{0}|= |T| \nonumber\\
\implies & |S_{0}|= \big[N(H):H\big]~\text{using (5)}\
\end{align}
From (2) ,
\begin{align}
& |S|\equiv |S_{0}|\big(mod~p \big) \nonumber\\
\implies & \big[G:H\big] \equiv \big[N(H):H\big]\big(mod~p \big) ~\text{using (3) and (6)}
\end{align}
2nd Part-
Let \(p\big| \big[G:H \big] \).
Then from (6),
&\begin{align*}
p\big| \big[N(H):H\big]\\
\implies & \big[N(H):H\big] \gt 1 ~\text{since p is a prime} \\
\implies & N(H)-H \ne \phi \\
\implies & N(H) \ne H \\
\end{align*}
Hence the theorem is proved.
Applications
Group Actions are crucial in a wide range of applications across mathematics and science. In geometry, group actions help classify shapes and structures based on their symmetries. In physics, they are used to study conservation laws and quantum mechanics. Group actions also play a role in coding theory, providing solutions to problems in communication systems. For further study, explore Relations and Ring Theory.
Conclusion
Burnside’s Theorem offers a streamlined approach to understanding symmetry and combinatorial enumeration. Its versatility in solving problems across multiple disciplines underscores its value in the study of Group Theory . Mastering this theorem enhances mathematical reasoning and problem-solving capabilities.
References
- Introduction to Group Theory by Benjamin Steinberg
- Topics in Group Theory by Geoffrey Smith
- Abstract Algebra by David S. Dummit and Richard M. Foote
- Algebra by Michael Artin
- Symmetry and Group Theory by Mark A. Armstrong
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FAQs
- What is Burnside’s Theorem?
Burnside’s Theorem is a method used in group actions to count distinct configurations of objects under symmetry. - Why is Burnside’s Theorem important?
It simplifies the process of counting distinct objects and helps solve problems involving symmetry and group actions. - What are the applications of Burnside’s Theorem?
This theorem is used in combinatorics, algebra, and computational mathematics. - How does Burnside’s Theorem work?
It calculates the average number of fixed points of a group acting on a set. - Is Burnside’s Theorem relevant to combinatorics?
Yes, it is an essential tool for counting problems involving symmetrical arrangements. - What is an example of Burnside’s Theorem?
Calculating the number of unique ways to color the faces of a cube is an example of applying Burnside’s Theorem. - What is the historical significance of Burnside’s Theorem?
Proposed by William Burnside, it marked a major advancement in group theory and combinatorics. - Can Burnside’s Theorem be applied in real-world scenarios?
Yes, it has applications in computer graphics, chemistry, and cryptography. - Where can I find exercises on Burnside’s Theorem?
Practice problems are available in Abstract Algebra Questions. - How is Burnside’s Theorem taught in universities?
It is part of advanced courses on Group Theory and abstract algebra.
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