Sylow's First Theorem

Sylow's First Theorem

Sylow's First Theorem provides profound insights into the existence of subgroups in finite groups. Originating in the works of Ludwig Sylow, this theorem has shaped much of Mathematics and Abstract Algebra. Its historical significance lies in its ability to identify and analyze Group Theory properties effectively. For further understanding, explore Mathematics problems or Abstract Algebra problems.

What You Will Learn?

  In this post, you will explore:

  • Sylow’s First Theorem: Let \((G,\circ)\) be a finite group of order \(p^{r}m\), where \(p\) be a prime and \(r,m\) be positive integers and \(p,m\) be relatively prime. Then \(G\) has subgroup of order \(p^{k}\), where \(k=0,1,2,…,r \).
  • Theorem-2: Let \((G,\circ)\) be a finite group and \(p\) be a prime. If \(p^{n}\big|~\big|G\big|\) then \(G\) has subgroup of order \(p^{n}\).

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 First Theorem is a cornerstone in Mathematics, particularly in Abstract Algebra. The theorem establishes criteria for identifying p-subgroups of finite groups. Its applications span diverse fields, including solving complex problems in Group Theory. Additional details can be found by exploring Mathematics problems.

Sylow's First Theorem

  Statement:
  Let \((G,\circ)\) be a finite group of order \(p^{r}m\), where \(p\) be a prime and \(r,m\) be positive integers and \(p,m\) be relatively prime. Then \(G\) has subgroup of order \(p^{k}\), where \(k=0,1,2,…,r \).


  Proof:
  Given that \((G,\circ)\) is a finite group of order \(n=p^{r}m\), where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime.
  To prove \(G\) has subgroups of order \(p^{k}\), where \(k=0,1,2,…,r \).
  We prove this theorem by using the principle of Mathematical Induction.

  • Let \(n=1\)
    Then \(r=0\), implies
    \(\set{e}\) is the required subgroup of order \(p^{k}\), where \(k=0 \).

  • Let the statement is true for all groups of order \(r\), where \(1\leq r \lt n\)
    Let \(G\) be finite group of order \(p^{r}m\), where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime.
    • Case-1: Let \(p\big|~\big|Z(G)\big| \)
      Then by Cauchy’s Theorem, \(\exists \) an element \(a\in Z(G)\) such that \(\big|a \big|=p \).
      Let us consider the subgroup \(H=\lang a\rang\).
      Then \(H\) is normal in \(G\).
      Implies \(G/H\) is exists and \begin{align*} \big|G/H\big| &=\frac{\big|G\big|}{\big|H\big|} \\ & =\frac{p^{r}m}{p} \\ & =p^{r-1}m \end{align*} Since \(p,m\) are relatively prime and \(\big|G/H\big|=p^{r-1}m\lt \big|G\big| \),
      then by hypothesis, \(G/H\) has subgroups of order \(p^{i},~~ i=0,1,2,…,r-1\).
      Let the subgroups of \(G/H\) are \(K_{i}/H\) such that \begin{align*} \big|K_{i}/H\big|=p^{i},~~ i=0,1,2,…,r-1 \end{align*} where \(K_{i}\) are subgroups of \(G\). Now
      \begin{align*} \big|K_{i}\big| & =\big|K_{i}/H\big|.\big|H\big|\\ & =p^{i}.p\\ & =p^{i+1},~~ i=0,1,2,…,r-1 \end{align*} Hence \(G\) has subgroups \(\set{e},K_{0},K_{1},…,K_{r-1}\) of order \(p^{0},p^{1},p^{2},….,p^{r} \).
    • Case-2: Let \(p\cancel{\big|}~\big|Z(G)\big| \)
      Since \(\big|G\big|=p^{r}m\) then \( p\big|~\big|G\big| \)
      then from the class equation \begin{align*} \big|G\big|=\big|Z(G)\big|+\displaystyle\sum_{a\notin Z(G)}^{}\big[G:C(a)\big] \end{align*} We claim that \(p\cancel{\big|}~\big[G:C(a)\big] \) where \(a\notin Z(G) \) since if \begin{align*} & p \big|~\big[G:C(a)\big]\\ \implies & p \big|~\big|Z(G)\big| ~\because p\big|~\big|G\big| \end{align*} A contradiction since \(p\cancel{\big|}~\big|Z(G)\big| \).
      Therefore \(\exists~~a\notin Z(G) \) such that \(p\cancel{\big|}~\big[G:C(a)\big] \).
      Now we have \begin{align*} & \big|G\big|=\big[G:C(a)\big].\big|C(a)\big|\\ \implies & p^{r} \big|~\big|C(a)\big| \because p\big|~\big|G\big| \text{ and } p\cancel{\big|}~\big[G:C(a)\big] \\ \implies & \big|C(a)\big|=p^{r}q \end{align*} for some positive integer \(q\) and \(p,q\) are relatively prime.
      Also since \(a\in G-Z(G)\) implies \(C(a)\ne G \) and \(a\ne e \).
      Therefore \(1 \lt C(a)\lt \big|G\big| \).
      Since \(p,q\) are relatively prime and \(\big|C(a)\big|=p^{r}q\lt \big|G\big| \),
      Then by hypothesis, \(C(a)\) has subgroups \(T_{i}\) of order \(p^{i}\), where \(i=0,1,2,…,r \).
      Since \(C(a)\) is a subgroup of \(G\),
      Hence \(G\) has subgroups \(T_{0},T_{1},…,T_{r}\) of order \(p^{0},p^{1},p^{2},….,p^{r} \).
    Therefore the statement is trues for \(\big|G\big|=n\).

  Then by the principle of mathematical induction, the statement is true for all \(n\).

Theorem-2

  Statement:
  Let \((G,\circ)\) be a finite group and \(p\) be a prime. If \(p^{n}\big|~\big|G\big|\) then \(G\) has subgroup of order \(p^{n}\).


  Proof:
  Given that \((G,\circ)\) is a finite group and \(p\) is a prime.
  Let \(p^{n}\big|~\big|G\big|\).
  To prove \(G\) has subgroup of order \(p^{n}\)
  Since \begin{align*} & p^{n}\big|~\big|G\big| \implies & \big|G\big|= p^{n}m \end{align*} for some positive integer \(m\) and \(p,m\) are relatively prime.
Then by Sylow’s theorem, \(G\) has subgroups \(K_{i},~~i=0,1,2,…,n \) of order \(p^{i},~~i=0,1,2,…,n \).
Hence \(K_{n}\) is the required subgroup of order \(p^{n}\).

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

  In conclusion, Sylow’s First Theorem is integral to understanding subgroup structures within Group Theory. Its significance in Mathematics and Abstract Algebra cannot be overstated. Explore related Mathematics problems and Abstract Algebra problems to deepen your comprehension.

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 First Theorem?
    Sylow’s First Theorem explains the existence of p-subgroups in finite groups, foundational to Group Theory.
  2. Why is Sylow’s First Theorem important?
    Sylow’s First Theorem is pivotal in Abstract Algebra, enabling subgroup analysis.
  3. Who proposed Sylow’s First Theorem?
    Sylow’s First Theorem was introduced by Ludwig Sylow, influencing Mathematics significantly.
  4. Where is Sylow’s First Theorem applied?
    Sylow’s First Theorem is used extensively in Abstract Algebra and Mathematics problems.
  5. How does Sylow’s First Theorem relate to Group Theory?
    Sylow’s First Theorem identifies critical subgroup structures within Group Theory.
  6. Is Sylow’s First Theorem part of modern Mathematics?
    Yes, Sylow’s First Theorem continues to impact Mathematics studies.
  7. What are p-subgroups in Sylow’s First Theorem?
    Sylow’s First Theorem defines p-subgroups as subsets of finite groups within Group Theory.
  8. How does Sylow’s First Theorem support problem-solving?
    Sylow’s First Theorem aids in solving Mathematics problems and Abstract Algebra problems.
  9. Can Sylow’s First Theorem be linked to other theorems?
    Yes, Sylow’s First Theorem complements results in Abstract Algebra.
  10. What prerequisites are needed to understand Sylow’s First Theorem?
    A strong foundation in Mathematics and Abstract Algebra is required.
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