Sylow's Second Theorem

Sylow's Second Theorem

Sylow's Second Theorem is a cornerstone of Abstract Algebra, specifically in Group Theory. Historically introduced to generalize group actions, this theorem ensures a deeper understanding of subgroup structures. Its practical applications include symmetry analysis, cryptography, and computational algebra. For more insights on related problems, visit Abstract Algebra Problems.

What You Will Learn?

  In this post, you will explore:

  • Sylow’s Second Theorem: Let \((G,\circ)\) be a finite group of order \(p^{r}m\), where \(p\) be a prime and \(r,m\) be two positive integers and \(p,m\) be relatively prime. Then any two Sylow p-subgroups of \(G\) are conjugate, then therefore isomorphic.
  • Theorem-2: Let \((G,\circ)\) be a finite group and \(H\) be a Sylow p-subgroup of \(G\) where \(p\) be a prime. Then \(H\) is a unique Sylow p-subgroup of \(G\) if and only if \(H\) is a normal subgroup of \(G\).

Things to Remember

Before diving into this post, make sure you are familiar with: Basic Definitions and Concepts of
  1. Set Theory
  2. Relations
  3. Mappings
  4. Group Theory

Introduction

  Sylow’s Second Theorem establishes the uniqueness of Sylow p-subgroups up to conjugacy. It serves as an essential tool in Abstract Algebra, linking fundamental properties of prime powers in group orders. Readers can delve deeper into Mathematics Notes to understand its role in modern algebra.

Sylow's Second Theorem

  Statement:
  Let \((G,\circ)\) be a finite group of order \(p^{r}m\), where \(p\) be a prime and \(r,m\) be two positive integers and \(p,m\) be relatively prime. Then any two Sylow p-subgroups of \(G\) are conjugate, then therefore isomorphic.


  Proof:
  Given that \((G,\circ)\) is a finite group of order \(p^{r}m\), where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime.
  To prove any two Sylow p-subgroups of \(G\) are conjugate.
  Let \(H\) and \(K\) are two Sylow p-subgroups of \(G\).
  Let \(S\) be the set of all lefts cosets of \(H\) in \(G\).
  Then \(\big|S\big|=\big[G:H\big] \)
  Since \(H\) is a Sylow p-subgroup of \(G\) then \(\big|G\big|=p^{r} \).
  Then we have \begin{align*} &\big|G\big|=\big[G:H\big].\big|H\big|\\ \implies & p^{r}m=\big[G:H\big].p^{r}\\ \implies & p ~\cancel{\big|}~ \big[G:H\big]\\ \implies & p ~\cancel{\big|}~ \big|S\big|~\because ~\big|S\big|=\big[G:H\big] \end{align*}   First we prove \(K\) acts on S.

  • Let us construct a mapping \(\cdot:K\times S\to S \) such that for all \(k\in K, a\in G,\) \begin{align*}k \cdot aH= (k\circ a)H \end{align*}
    • To prove \(\cdot \) is well defined.
      Let \(k_{1},k_{2}\in K\) and \(aH,bH\in S\) where \(a,b\in G\) such that \begin{align*} & \big(k_{1},aH \big)=\big(k_{2},bH \big)\\ \implies & k_{1}=k_{2},~ aH=bH\\ \implies & k_{1}=k_{2},~ b^{-1}\circ a\in H\\ \implies & b^{-1}\circ k^{-1}_{2}\circ k_{1}\circ a\in H\\ \implies & \big(k_{2}\circ b\big)^{-1}\circ\big(k_{1}\circ a\big)\in H\\ \implies & \big(k_{1}\circ a\big)H=\big(k_{2}\circ b\big)H\\ \implies & k_{1} \cdot aH =k_{2} \cdot bH \end{align*} Therefore \(\cdot \) is well defined.
    • Let \(k_{1},k_{2}\in K\) and \(aH\in S\) where \(a\in G\). Now \begin{align*} \big(k_{1}\circ k_{2}\big) \cdot aH & = \big(\big(k_{1}\circ k_{2}\big)\circ a\big)H\\ & = \big(k_{1}\circ \big( k_{2}\circ a\big)\big)H \\ & = k_{1}\cdot \big( k_{2}\circ a\big)H \\ & = k_{1}\cdot \big( k_{2}\cdot aH\big) \\ \end{align*}
    • Let \( e\) be the identity element of \(G\) and \(aH\in S\) where \(a\in G\). Now \begin{align*} e \cdot aH= \big( e\circ a\big)H = aH \end{align*}

  Therefore \(K\) acts on S.
  Let \(S_{0}=\set{aH\in S: k \cdot aH= aH ~\forall~ k\in K} \).
  Since \(K\) is a Sylow p-subgroup of \(G\) then \(\big|K\big|=p^{r} \). Therefore \begin{align*} & \big|S\big|=\big|S_{0}\big|\big(~mod~p\big) \\ \implies & p ~\cancel{\big|} \big|S_{0}\big|~\text{ and }~\big|S_{0}\big|\ne 0~\because~p ~\cancel{\big|}~ \big|S\big| \\ \implies & S_{0} \ne \phi \end{align*}    Let \(\alpha H\in S_{0}\) where \(\alpha \in G\). Then we have \begin{align*} & k\cdot \alpha H =\alpha H~\forall~k\in K \\ \implies & \big(k\circ \alpha \big) H =\alpha H~\forall~k\in K \\ \implies & \alpha^{-1}\circ\big(k\circ \alpha \big) \in H~\forall~k\in K \\ \implies & \big(\alpha^{-1} K \alpha \big) \subseteq H \end{align*}   Since \(H\) and \(K\) are Sylow p-subgroups of \(G\) then \(\big|H\big|=\big|K\big| \). Therefore \begin{align*} & \big|\alpha^{-1} K \alpha\big|=\big|K\big| \\ \implies & \big|\alpha^{-1} K \alpha\big|=\big|H\big|\\ \implies & \alpha^{-1} K \alpha=H~\because~\big(\alpha^{-1} K \alpha \big) \subseteq H \end{align*}   Therefore \(H\) and \(K\) are conjugates.
  Since \(H\) and \(K\) are any two Sylow p-subgroups of \(G\) then any two Sylow p-subgroups of \(G\) are conjugates.

Theorem-2

  Statement:
  Let \((G,\circ)\) be a finite group and \(H\) be a Sylow p-subgroup of \(G\) where \(p\) be a prime. Then \(H\) is a unique Sylow p-subgroup of \(G\) if and only if \(H\) is a normal subgroup of \(G\).


  Proof:
  Given that \((G,\circ)\) is a finite group and \(H\) is a Sylow p-subgroup of \(G\) where \(p\) is a prime.
  Let \(\big|G\big|=p^{r}m\) where \(r,m\) are positive integers and \(p,m\) are relatively prime.

  • Let \(H\) be a unique Sylow p-subgroup of \(G\).
    To prove \(H\) is a normal subgroup of \(G\).
    Let \(\alpha\in G \)
    Implies \(\alpha H \alpha^{-1}\) is a Sylow p-subgroup of \(G\).
    But \(H\) is the unique Sylow p-subgroup of \(G\).
    Then \(\alpha H \alpha^{-1}=H\)
    Since \(\alpha\in G \) is an arbitrary element then \(H\) is a normal subgroup of \(G\).
  • Let \(H\) be a normal subgroup of \(G\).
    To prove \(H\) is a unique Sylow p-subgroup of \(G\).
    We have \(\alpha H \alpha^{-1}=H~\forall~\alpha\in G\).
    Let \(K\) be a Sylow p-subgroup of \(G\).
    Then by Sylow’s Second theorem, \(H\) and \(K\) are conjugates.
    Then \(\exists \) a \(\beta\in G\) such that \begin{align*} & \alpha H \alpha^{-1}=K \\ \implies & H=K ~\because~\alpha H \alpha^{-1}=H~\forall~\alpha\in G \end{align*} Hence \(H\) is a unique Sylow p-subgroup of \(G\).

Conclusion

  Sylow’s Second Theorem remains an indispensable part of Group Theory, providing insights into subgroup behavior under specific conditions. Its applications have been extended across various fields, reaffirming its significance in theoretical and applied Mathematics. Explore related Abstract Algebra Questions for practical problems.

References

  1. Introduction to Group Theory by Benjamin Steinberg
  2. Topics in Group Theory by Geoffrey Smith
  3. Abstract Algebra by David S. Dummit and Richard M. Foote
  4. Algebra by Michael Artin
  5. Symmetry and Group Theory by Mark A. Armstrong

FAQs

  1. What is Sylow’s Second Theorem?
    Sylow’s Second Theorem ensures the uniqueness of Sylow p-subgroups up to conjugacy and plays a key role in Abstract Algebra.
  2. Why is Sylow’s Second Theorem important?
    It provides insights into subgroup structures, which are vital in Mathematics and related fields.
  3. How does Sylow’s Second Theorem apply in Group Theory?
    It classifies p-subgroups based on prime factorization of group orders. Learn more in Group Theory.
  4. What are Sylow p-subgroups?
    Subgroups of order \(p^{n} \) in a finite group, where \(p \) is a prime divisor of the group’s order.
  5. Where can I find problems on Sylow’s Second Theorem?
    Visit Abstract Algebra Problems for practice questions.
  6. What is the relationship between Sylow’s First and Second Theorems?
    The first theorem identifies existence, while the second ensures uniqueness up to conjugacy.
  7. Are there real-world applications of Sylow’s Second Theorem?
    Yes, including applications in cryptography and computational algebra.
  8. Can Sylow’s Second Theorem be extended to infinite groups?
    It applies primarily to finite groups, but analogous results exist for infinite cases.
  9. What are conjugacy classes in Group Theory?
    Sets of elements that are conjugate to each other; explore more in Group Theory.
  10. How do Sylow Theorems help in understanding group structures?
    They decompose groups into manageable subgroups, offering insight into their composition.
Scroll to Top