Cauchy's Theorem
Cauchy's Theorem
Cauchy's Theorem is a foundational result in Abstract Algebra. Historically, it provided a critical understanding of group theory by demonstrating the existence of elements of specific orders in finite groups. The importance of Cauchy's Theorem lies in its role in analyzing group structures, making it a cornerstone in modern mathematical studies.
What You Will Learn?
In this post, you will explore:
- Theorem-1: If \((G,\circ)\) is a finite commutative group of order \(n\) such that \(p\big| n\) where \(p\) is a prime then \(G\) contains an element of order \(p\).
- Cauchy’s Theorem: If \((G,\circ)\) is a finite group of order \(n\) such that \(p\big| n\) where \(p\) is a prime then \(G\) contains an element of order \(n\) and \(G\) has a subgroup of order \(p\).
Things to Remember
- Set Theory
- Relations
- Mappings
- Group Theory
Introduction
Cauchy’s Theorem is a pivotal topic in Mathematics. This theorem states that for any prime number dividing the order of a finite group, there exists an element of that group whose order is the prime. It plays a significant role in Abstract Algebra, forming the basis for advanced group theory concepts.
Theorem-1
Statement:
If \((G,\circ)\) is a finite commutative group of order \(n\) such that \(p\big| n\) where \(p\) is a prime then \(G\) contains an element of order \(n\).
Proof:
Let \((G,\circ)\) be a finite commutative group of order \(n\) such that \(p\big| n\) where \(p\) be a prime.
To prove \(\exists\) an element in \(G\) of order \(p\)
\(\because p\big|n\implies n\geq 2\).
We prove this theorem by the principle of Mathematical Induction.
- Let \(n=2\)
\(\implies \big|G\big|=2\)
\(\implies p=2\)
Since \(\big|G\big|\) is a prime then \(G\) is cyclic.
Then the element in \(G\), other than identity element \(e\), is of order \(2\).
- Let \(n\) is a prime
\(\implies \big|G\big|\) is prime
Then every element in \(G\), other than identity element \(e\), is of order \(p\).
- Let the statement is true for all groups of order \(r\), where \(2\leq r \lt n\)
Let \(G\) be a group of order \(n\) such that \(p\big|n\) where \(p\) is prime.
Let \(a\in G\) where \(a\ne e\) such that \(\big|a\big|=m\).
- Case-1: Let \(p\big|m\)
\begin{align*} & m=pk ~\text{ for some positive integer } k \\ \implies & a^{pk}=e \\ \implies & \big(a^{k}\big)^{p}=e \\ \implies & \big|a^{k}\big|=p~\big[\because ~p\text{ is a prime and }a\ne e \big] \\ \end{align*} \(\therefore a^{k}\) is the required element of order \(p\).
- Case-2: Let \(p\cancel{\big|}m\)
Let \(H=\lang a \rang\) then \(H\) is cyclic subgroup such that \(\big|H\big|=m\).
Since \(G\) is commutative then \(H\) is normal in \(G\) implies the quotient group \(G/H\) exists. Then \begin{align*} & \big|G\big|=\big|G/H\big|.\big|H\big|\\ \implies & p \big|~ \big|G/H\big|~\big[\because p \big|~\big|G\big| \text{ and } p\cancel{\big|}~\big|H\big| \big] \end{align*} \(\therefore G/H\) is a group of order less than \(n\) and \(p \big|~ \big|G/H\big| \).
Then by hypothesis, \(\exists\) an element in \(G/H\) of order \(p\).
Let \(bH\in G/H \) such that \begin{align*} & \big|bH\big|=p \\ \implies & \big(bH \big)^{p}=H \\ \implies & b^{p}H =H \\ \implies & b^{p} \in H \\ \implies & \big(b^{p})^{m}=e ~\big[\because H \text{ a is cyclic subgroup of order }m \big]\\ \implies & \big(b^{m})^{p}=e\\ \end{align*} If possible let \(b^{m}=e\) then \begin{align*} & b^{m}H =H \\ \implies & \big(bH \big)^{m}=H \\ \implies & p\big| m ~\big[\because \big|bH\big|=p \big]\\ \end{align*} A contradiction since \(p\cancel{\big|}m \).
\(\therefore b^{m}\ne e \). But \(\big(b^{m})^{p}=e\) and \(p\) is a prime.
Therefore \(\big|b^{m}\big|=p\).
Implies \( b^{m}\) is the required element of order \(p\).
- Case-1: Let \(p\big|m\)
Therefore the statement is true for \(n\).
Then by the principle of mathematical induction, the statement is true for all \(n\geq 2\).
Hence theorem is proved.
Cauchy's Theorem
Statement:
If \((G,\circ)\) is a finite group of order \(n\) such that \(p\big| n\) where \(p\) is a prime then \(G\) contains an element of order \(n\) and \(G\) has a subgroup of order \(p\).
Proof:
Let \((G,\circ)\) be a finite group of order \(n\) such that \(p\big| n\) where \(p\) be a prime.
To prove \(\exists\) an element in \(G\) of order \(p\) and \(G\) has a subgroup of order \(p\)
\(\because p\big|n\implies n\geq 2\).
We prove this theorem by the principle of Mathematical Induction.
- Let \(n=2\)
\(\implies \big|G\big|=2\)
\(\implies p=2\)
Since \(\big|G\big|\) is a prime then \(G\) is cyclic.
Then the element in \(G\), other than identity element \(e\), is of order \(2\).
- Let \(n\) is a prime
\(\implies \big|G\big|\) is prime
Then every element in \(G\), other than identity element \(e\), is of order \(p\).
- Let the statement is true for all groups of order \(r\), where \(2\leq r \lt n\)
Let \(G\) be a group of order \(n\) such that \(p\big|n\) where \(p\) is prime.
- Case-1: Let \(Z(G)=G\)
Then \(G\) is commutative then \(\exists\) an element in \(G\) of order \(p\).
- Case-2: Let \(Z(G)\ne G\)
Then \(Z(G)\subset G \implies G-Z(G)\ne \phi\) and \(\big|Z(G)\big|\lt \big|G\big|\).
Let \begin{align*} & a\in G-Z(G) \\ \implies & a\in G \text{ and } a \notin Z(G) \\ \implies & C(a)\ne G \\ \implies & \big[G:C(a)\big] \gt 1 \\ \implies & 1\lt \big|C(a)\big|\lt \big|G \big|~ \big[\because a\ne e \text{ and } a\in C(a) \big] \end{align*}- Subcase-1: Let \(p\big|~\big|C(a)\big| \)
Since \(1\lt \big|C(a)\big|\lt \big|G \big| \), then hypothesis, \(\exists\) an element in \(C(a)\) of order \(p\).
Therefore \(\exists\) an element in \(G\) of order \(p\) since \(C(a)\subset G \).
- Subcase-2: Let \(p\cancel{\big|}~\big|C(a)\big| \)
Now from Lagrange’s theorem, \begin{align*} & \big|G\big|=\big[G:C(a)\big].\big|C(a)\big| \\ \implies & p\big|~\big[G:C(a)\big]~ \because ~p\big|~\big|G\big| \text{ and }p\cancel{\big|}~\big|C(a)\big|\\ \implies & p\big|~\displaystyle\sum_{a\notin Z(G)}^{}\big[G:C(a)\big] \end{align*} Also from the Class Equation , \begin{align*} & \big|G\big|=\big|Z(G)\big|+\displaystyle\sum_{a\notin Z(G)}^{}\big[G:C(a)\big]\\ \implies & p\big|~\big|Z(G)\big| \because ~p\big|~\big|G\big|\text{ and }p\big|~\displaystyle\sum_{a\notin Z(G)}^{}\big[G:C(a)\big]\\ \implies &\big|Z(G)\big|\gt 1 \end{align*} Therefore we have \(1\lt \big|Z(G)\big| \lt \big|G\big| \) and \( p\big|~\big|Z(G)\big|\). Then by hypothesis, \(\exists\) an element in \(Z(G)\) of order \(p\).
Therefore \(\exists\) an element in \(G\) of order \(p\) since \(Z(G)\subset G \).
- Subcase-1: Let \(p\big|~\big|C(a)\big| \)
- Case-1: Let \(Z(G)=G\)
Therefore the statement is true for \(n\).
Then by the principle of mathematical induction, the statement is true for all \(n\geq 2\).
Let the element \(\alpha \in G\) such that \(\big|\alpha \big|=p\).
Then \(H=\lang\alpha\rang \) is the required subgroup of \(G\) such that \(\big|H\big|=p\).
Hence theorem is proved.
Theorem-3
Statement:
If \((G,\circ)\) is a finite commutative group of order \(n\) such that \(m\big| n\) where \(m\) is a positive integer then \(G\) has a subgroup of order \(m\).
Proof:
Let \((G,\circ)\) be a finite commutative group of order \(n\) such that \(m\big| n\) where \(m\) be a positive integer.
To prove \(G\) has a subgroup of order \(m\)
We prove this theorem by the principle of Mathematical Induction.
- Let \(n=1\).
Then \(m=1\) implies \(\set{e} \) is the required subgroup of order \(m\).
- Let \(n=2\).
Then \(m=2\) implies \(\set{G} \) is the required subgroup of order \(m\).
- Let the statement is true for all groups of order \(r\), where \(2\leq r \lt n\)
Let \((G,\circ)\) be a finite commutative group of order \(n\) such that \(m\big| n\) where \(m\) be a positive integer.
Let \(n=mq\) for some positive integer \(q\).
Also \(\exists\) a prime number \(p\) such that \begin{align*} & p\big|~m \\ \implies & m=pk ~\text{ for some positive integer }k\\ \implies & p\big|~n \big[\because m\big|~n \big] \end{align*} Then by Cauchy’s theorem, \(\exists\) an element in \(G\) of order \(p\).
Let the element \(\alpha \in G\) such that \(\big|\alpha \big|=p\).
Then \(H=\lang\alpha\rang \) is the cyclic subgroup of \(G\) such that \(\big|H\big|=p\).
Since \(G\) is commutative then \(H\) is normal in \(G\). Implies the quotient group \(G/H \) exists.
Now we have \begin{align*} & \big|G/H \big|= \frac{\big|G \big|}{\big|H \big|}\\ \implies & \big|G/H \big|= \frac{n}{p} \\ \implies & \big|G/H \big|= \frac{mq}{p} ~\big[\because n=mq \big]\\ \implies & \big|G/H \big|= \frac{pkq}{p}~\big[\because m=pk \big] \\ \implies & \big|G/H\big|= kq \\ \implies & k\big|~ \big|G/H \big|\\ \end{align*} Therefore \(1\leq \big|~ \big|G/H \big| \lt \big|G \big|\) and \(k\big|~ \big|G/H \big| \).
Then by induction hypothesis, \(G/H\) has a subgroup \(K/H\) of order \(k\) where \(K\) is a subgroup of \(G\) and \(H\) is normal in \(K\). Now \begin{align*} & \big|K/H \big|=\frac{\big|K \big|}{\big|H \big|}\\ \implies & \big|K \big|=\big|K/H \big|.\big|H \big|\\ \implies & \big|K \big|=kp\\ \implies & \big|K \big|=m \end{align*} Therefore \(K\) is the required subgroup of \(G\) of order \(m\)
Therefore the statement is true for \(n\).
Then by the principle of mathematical induction, the statement is true for all \(n\).
Hence 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
Cauchy’s Theorem underpins many results in Abstract Algebra. Its ability to ensure the existence of specific elements in finite groups highlights its mathematical elegance and utility, making it indispensable in the study of Mathematics.
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 Cauchy’s Theorem?
It is a theorem in group theory stating that if a prime number divides the order of a finite group, there exists an element of that group whose order is the prime. - Why is Cauchy’s Theorem important?
It helps in understanding the structure of finite groups and the existence of elements of specific orders. - Where is Cauchy’s Theorem applied?
It is widely applied in Abstract Algebra and other areas of mathematics. - How is Cauchy’s Theorem proved?
The proof involves showing the existence of a subgroup generated by an element of the given order using group actions and counting arguments. - What are related topics to Cauchy’s Theorem?
Related topics include Sylow Theorems, Lagrange’s Theorem, and Abstract Algebra. - Can Cauchy’s Theorem be applied to infinite groups?
No, it is specifically applicable to finite groups. - What is the historical significance of Cauchy’s Theorem?
It was one of the early results that laid the foundation for modern group theory. - How does Cauchy’s Theorem relate to Sylow Theorems?
Cauchy’s Theorem is a special case of the Sylow Theorems for groups of prime order. - Where can I find practice questions on Cauchy’s Theorem?
Practice questions are available at Mathematics Questions and Abstract Algebra Questions. - What is the connection between Cauchy’s Theorem and Lagrange’s Theorem?
Cauchy’s Theorem builds on Lagrange’s Theorem by identifying elements of specific orders when the group’s order is divisible by a prime.
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