Simplicity of \(A_{n}\)
Simplicity of \(A_{n}\)
Simplicity of \(A_{n}\) has played a crucial role in the development of Mathematics, especially in Abstract Algebra. This concept establishes the properties of alternating groups, an essential part of Group Theory. Historically, the proof of simplicity for alternating groups \(A_{n},~n \geq 5\) was a major milestone, showing the robustness of group structures. Further explorations can be found in Alternating Groups.
What You Will Learn?
In this post, you will explore:
- Definition: Simple Group
- Theorem-1: Let \(H\) be a normal subgroup of \(A_{n},~n\geq 5\). If \(H\) contains a 3-cycle then \(H=A_{n}\).
- Theorem-2: Let \(H\) be a normal subgroup of \(A_{n},~n\geq 5\). If \(H\) contains a product of two disjoint transpositions then \(H=A_{n}\).
- Theorem-3: If \(n\geq 5 \) then \(A_{n}\) is simple.
Things to Remember
- Set Theory
- Relations
- Mappings
- Group Theory
Introduction
The Simplicity of \(A_{n}\) theorem highlights the indivisibility of alternating groups \(A_{n}\) for \(n \geq 5\). It is a foundational topic in Mathematics, forming the backbone of studies in Abstract Algebra. These groups are vital for understanding the symmetry and structure of algebraic systems, making their study indispensable for learners of Group Theory. Problems based on this theorem are available here.
Simple Group
Definition:
Let \((G,\circ)\) be a group. Then \(G\) is said to be a Simple Group if \(G\) is non-trivial and the only normal subgroups of \(G\) are \(\set{e} \) and \(G\).
Theorem-1
Statement:
Let \(H\) be a normal subgroup of \(A_{n},~n\geq 5\). If \(H\) contains a 3-cycle then \(H=A_{n}\).
Proof:
Given that \(H\) is a normal subgroup of \(A_{n},~n\geq 5\).
Let \(H\) contains a 3-cycle \(\big(a,b,c\big)\).
To prove \(H=A_{n}\).
That is to prove \(A_{n} \subseteq H\) since \(H \subseteq A_{n}\).
Let \(\big(u,v,w\big)\in A_{n}\).
Let \(\pi\in S_{n}\) such that
\begin{align*}
\pi(a)=u,~\pi(b)=v,~\pi(c)=w
\end{align*}
Therefore
\begin{align*}
& \pi\circ \big(a,b,c\big)\circ \pi^{-1}=\big(\pi(a),\pi(b),\pi(c)\big)\\
\implies & \pi\circ \big(a,b,c\big)\circ \pi^{-1}=\big(u,v,w\big)
\end{align*}
- Case-1: Let \(\pi\in A_{n}\)
Since \(H\) is normal in \(A_{n}\) then \begin{align*} & \pi\circ \big(a,b,c\big)\circ \pi^{-1} \in H \\ \implies & \big(u,v,w\big) \in H \\ \implies & A_{n} \subseteq H \\ \implies & H=A_{n} \end{align*} - Case-2: Let \(\pi\notin A_{n}\)
Then \(\pi \) is an odd permutation.
Since \(n\geq 5\) then \(\exists\) two members \(d,f\in I_{n}\) such that they are distinct from \(a,b\) and \(c\).
\(\therefore \pi\circ \big(d,f \big) \in A_{n}\). Now \begin{align*} & \big(u,v,w\big) \\ & = \pi\circ \big(a,b,c\big)\circ \pi^{-1} \\ & = \pi\circ \big(a,b,c\big)\circ \big(d,f\big)\circ \big(d,f\big)^{-1} \circ \pi^{-1} \\ & = \pi\circ \big(d,f\big)\circ \big(a,b,c\big)\circ \big(d,f\big)^{-1} \circ \pi^{-1}~\because H \text{ is normal in } A_{n} \\ & = \big(\pi\circ \big(d,f\big) \big)\circ \big(a,b,c\big)\circ \big(\pi\circ \big(d,f\big) \big)^{-1}\\ & \in H ~\because H \text{ is normal in } A_{n} \end{align*} Therefore \begin{align*} & \big(u,v,w\big) \in H \\ \implies & A_{n} \subseteq H \\ \implies & H=A_{n} \end{align*}
Hence the theorem in proved.
Theorem-2
Statement:
Let \(H\) be a normal subgroup of \(A_{n},~n\geq 5\). If \(H\) contains a product of two disjoint transpositions then \(H=A_{n}\).
Proof:
Given that \(H\) is a normal subgroup of \(A_{n},~n\geq 5\).
Let \(H\) contains a product of two disjoint transpositions \( \big(a,b \big)\circ \big(c,d \big) \in H \) where \( \big(a,b \big)\) and \( \big(c,d \big)\) are two disjoint transpositions.
To prove \(H=A_{n}\).
That is to prove \(A_{n} \subseteq H\) since \(H \subseteq A_{n}\).
Let \(\pi=\big(c,d,w\big)\in A_{n}\).
Since \(n\geq 5\) then \(\exists\) one member \(w\in I_{n}\) such that they are distinct from \(a,b,c\) and \(d\).
Since \(H\) is normal in \(A_{n} \) then
\begin{align*}
& \pi \circ \big(a,b \big)\circ \big(c,d \big) \circ \pi^{-1} \in H \\
\implies & \big(c,d,w\big) \circ \big(a,b \big)\circ \big(c,d \big) \circ \big(c,d,w\big)^{-1} \in H \\
\implies & \big(c,d,w\big) \circ \big(a,b \big)\circ \big(c,d \big) \circ \big(c,w,d\big)\in H \\
\implies & \big(c,d,w\big) \circ \big(a,b \big)\circ \big(c,w \big)\in H \\
\implies & \big(c,d,w\big) \circ \big(c,w \big) \circ \big(a,b \big)\in H \\
\implies & \big(d,w\big) \circ \big(a,b \big)\in H \\
\implies & \big(a,b \big)\circ \big(c,d \big) \circ \big(d,w\big) \circ \big(a,b \big)\in H~ \because \big(a,b \big)\circ \big(c,d \big) \in H \\
\implies & \big(a,b \big)\circ \big(c,d, w \big) \circ \big(a,b \big)\in H \\
\implies & \big(a,b \big) \circ \big(a,b \big) \circ \big(c,d, w \big) \in H \\
\implies & \big(c,d, w \big) \in H
\end{align*}
Therefore \(H\) contains a 3-cycle.
Hence \(H=A_{n}\).
Theorem-3
Statement:
If \(n\geq 5 \) then \(A_{n}\) is simple.
Proof:
Let \(n\geq 5 \).
To prove \(A_{n}\) is simple.
Let \(H\) is a normal subgroup of \(A_{n}\) and \(H\ne \set{e}\).
Let \(\pi \in H\) and \(\pi \ne e\) be a permutation that moves the smallest number of elements, say \(m\).
Then \(m \geq 3\).
If possible let \(m \gt 3\).
Let \(\pi=\pi_{1}\circ\pi_{2}\circ … \circ \pi_{k} \) where \(\pi_{i},~i=1,2,…,k \) are disjoint cycles.
If possible let \(\pi_{i},~i=1,2,…,k \) are all transpositions.
Then clearly \(k \geq 2\).
Let \(\pi_{1}=\big(a,b \big)\) and \(\pi_{2}=\big(c,d \big)\) and \(f\in I_{n} \) such that \(f\notin \set{a,b,c,d} \).
Let \(\sigma=\big(c,d,f \big)\in A_{n}\).
Since \(H\) is normal in \(A_{n}\) then
\begin{align*}
& \sigma \circ \pi \circ \sigma^{-1} \in H \\
\implies & \pi^{-1} \circ \sigma \circ \pi \circ \sigma^{-1} \in H
\end{align*}
Let \(\pi^{\prime}= \pi^{-1} \circ \sigma \circ \pi \circ \sigma^{-1}\).
Let \(u\in I_{n} \) such that \(u\notin \set{a,b,c,d,f} \) such that \(\pi(u)=u \). Then
\begin{align*}
\pi^{\prime}(u) &=\big(\pi^{-1} \circ \sigma \circ \pi \circ \sigma^{-1} \big)(u) \\
& =\pi^{-1}\big(\sigma \big(\pi \big(\sigma^{-1}(u) \big) \big) \big)\\
& =\pi^{-1}\big(\sigma \big(\pi (u) \big) \big)\\
& =\pi^{-1}\big(\sigma (u) \big)\\
& =\pi^{-1} (u) \\
& =u
\end{align*}
and
\begin{align*}
\pi^{\prime}(a) &=\big(\pi^{-1} \circ \sigma \circ \pi \circ \sigma^{-1} \big)(a)\\
& =\pi^{-1}\big(\sigma \big(\pi \big(\sigma^{-1}(a) \big) \big) \big)\\
& =\pi^{-1}\big(\sigma \big(\pi (a) \big) \big)\\
& =\pi^{-1}\big(\sigma (b) \big)\\
& =\pi^{-1} (b) \\
& =a
\end{align*}
and
\begin{align*}
\pi^{\prime}(b) &=\big(\pi^{-1} \circ \sigma \circ \pi \circ \sigma^{-1} \big)(b) \\
& =\pi^{-1}\big(\sigma \big(\pi \big(\sigma^{-1}(b) \big) \big) \big)\\
& =\pi^{-1}\big(\sigma \big(\pi (b) \big) \big)\\
& =\pi^{-1}\big(\sigma (a) \big)\\
& =\pi^{-1} (a) \\
& =b
\end{align*}
but
\begin{align*}
\pi^{\prime}(f) &=\big(\pi^{-1} \circ \sigma \circ \pi \circ \sigma^{-1} \big)(f)\\
& =\pi^{-1}\big(\sigma \big(\pi \big(\sigma^{-1}(f) \big) \big) \big)\\
& =\pi^{-1}\big(\sigma \big(\pi (d) \big) \big)\\
& =\pi^{-1}\big(\sigma (c) \big)\\
& =\pi^{-1} (d) \\
& =c
\end{align*}
Therefore \(\pi^{\prime}\ne e \) and \(\pi^{\prime} \in H \) but \(\pi^{\prime} \) moves fewer elements than \(\pi\).
A contradiction since \(\pi\) moves smallest number of elements.
Therefore our assumption is wrong.
Hence \(\exists\) a permutation \(\pi_{i}\), \(1\geq i \geq k\) such that the length of the cycle of \( \pi_{i}\) is greater than or equal to \(3\).
Without loss of generality, let \( \pi_{1}=\big( a,b,c,…\big)\).
- Let \(m=4\)
Then \(\pi\) is a cycle of length \(4\).
\(\pi\) is an odd permutation.
A contradiction since \(\pi\in A_{n}\).
\(\therefore m\ne 4\) \(\). - Let \(m\geq 5\)
Then \(\pi\) moves at least \(5\) elements.
Since \(n \geq 5 \) then \(\exists ~g,h\in I_{n}\) such that \(g,h\notin \set{a,b,c}\).
Let \(\tau=\big(c,g,h \big)\in A_{n} \).
Since \(H\) is normal in \(A_{n}\) then \begin{align*} & \tau \circ \pi \circ \tau^{-1} \in H \\ \implies & \pi^{-1} \circ \tau \circ \pi \circ \tau^{-1} \in H \end{align*} Let \(\pi^{\prime\prime}= \pi^{-1} \circ \tau \circ \pi \circ \tau^{-1}\).
Let \(w\in I_{n} \) such that \(w\notin \set{a,b,c,g,h} \) such that \(\pi(w)=w \). Then \begin{align*} \pi^{\prime\prime}(w) &=\big(\pi^{-1} \circ \tau \circ \pi \circ \tau^{-1} \big)(w) \\ & =\pi^{-1}\big(\tau \big(\pi \big(\tau^{-1}(w) \big) \big) \big)\\ & =\pi^{-1}\big(\tau \big(\pi (w) \big) \big)\\ & =\pi^{-1}\big(\tau (w) \big)\\ & =\pi^{-1} (w) \\ & =w \end{align*} and \begin{align*} \pi^{\prime\prime}(a) &=\big(\pi^{-1} \circ \tau \circ \pi \circ \tau^{-1} \big)(a) \\ & =\pi^{-1}\big(\tau \big(\pi \big(\tau^{-1}(a) \big) \big) \big)\\ & =\pi^{-1}\big(\tau \big(\pi (a) \big) \big)\\ & =\pi^{-1}\big(\tau (b) \big)\\ & =\pi^{-1} (b) \\ & =a \end{align*} but \begin{align*} \pi^{\prime\prime}(b) &=\big(\pi^{-1} \circ \tau \circ \pi \circ \tau^{-1} \big)(b) \\ & =\pi^{-1}\big(\tau \big(\pi \big(\tau^{-1}(b) \big) \big) \big)\\ & =\pi^{-1}\big(\tau \big(\pi (b) \big) \big)\\ & =\pi^{-1}\big(\tau (c) \big)\\ & =\pi^{-1} (g) \\ & \ne b ~\because \pi^{-1}(c)=b \end{align*} Therefore \(\pi^{\prime\prime}\ne e \) and \(\pi^{\prime\prime} \in H \) but \(\pi^{\prime\prime} \) moves fewer elements than \(\pi\).
A contradiction since \(\pi\) moves smallest number of elements.
\(\therefore m\ngeq 5\).
\(\therefore m= 3\).
Implies \(H\) contains a 3-cycle.
\( \therefore H=A_{n}\).
Implies \(A_{n}\) has only two normal subgroups \(\set{e} \) and \(G\).
Hence \(A_{n}\) is simple.
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
The Simplicity of \(A_{n}\) continues to be a topic of great importance in Mathematics. Its applications in Abstract Algebra and Group Theory underscore its significance in modern mathematical theory. For deeper insights and advanced problems, visit our Mathematics Questions section.
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 the Simplicity of \(A_{n}\)?
The Simplicity of An refers to the property of alternating groups An (n ≥ 5) being simple groups, studied in Abstract Algebra. - Why is Simplicity of \(A_{n}\) important?
The Simplicity of \(A_{n}\) is important for understanding group structures in Group Theory. - Which groups are simple?
Alternating groups \(A_{n},~n\geq 5\) are proven simple under the Simplicity of An theorem. - How is Simplicity of \(A_{n}\) proven?
The proof involves techniques from Group Theory and relies on permutation properties. - Where can I find problems on Simplicity of \(A_{n}\)?
Problems on the Simplicity of \(A_{n}\) can be found here. - Who discovered the Simplicity of \(A_{n}\)?
Historical contributions to the Simplicity of \(A_{n}\) were made by Évariste Galois and others in Abstract Algebra. - What are the applications of Simplicity of \(A_{n}\)?
Applications of the Simplicity of \(A_{n}\) include studies in Mathematics and symmetry. - What is the smallest simple group?
The smallest simple group is \(A_{5}\), studied under the Simplicity of \(A_{n}\) theorem. - What is the significance of alternating groups?
Alternating groups, proven simple for \(n\geq 5\), are essential in Group Theory. - Are all An groups simple?
No, the Simplicity of \(A_{n}\) applies only to alternating groups \(A_{n}\) for \(n\geq 5\).
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