Normalizer of a Subgroup - Definition and Key Theorems

Normalizer of a Subgroup

Normalizer of a Subgroup is a critical concept in Abstract Algebra. Historically, it was introduced as a part of group theory to explore the structural relationships within groups. The importance of the Normalizer of a Subgroup lies in its ability to identify elements that preserve subgroup structure under conjugation, thus aiding in advanced mathematical studies and research.

What You Will Learn?

  In this post, you will explore:

  • Definition: Invariant under an Element
  • Definition: Normalizer of a Subgroup
  • Theorem-1: Let \((G,\circ)\) be a group and \(H\) and \(K\) be two subgroup of \(G\). Then the normalizer \(N_{K}(H) \) of \(H\) in \(K\) is subgroup of \(K\).
  • Theorem-2: Let \((G,\circ)\) be a group and \(H\) and \(K\) be two subgroup of \(G\). Then the number of distinct conjugates of \(H\) induced by the elements of \(K\) is equal to \(\big[K:N_{K}(H) \big]\).
  • Theorem-3: Let \((G,\circ)\) be a group and \(H\) and \(K\) be two finite subgroups of \(G\). If \(H\) is invariant of under \(n\) elements of \(K\), then \(H\) has \(\frac{\big|K\big|}{n}\) conjugates by elements of \(K\).

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

  Normalizer of a Subgroup is an essential topic in Mathematics. It refers to the set of all elements in a group that commute with every element of a given subgroup under conjugation. This concept finds applications in Abstract Algebra and is widely studied across various university curricula.

Invariant under an Element

  Definition:
  Let \((G,\circ)\) be a group and \(H\) be a subgroup of \(G\) and \(a\in G\). Then \(H\) is sais to be invariant under \(a\) if \(aHa^{-1}=H \).

Normalizer of a Subgroup

  Definition:
  Let \((G,\circ)\) be a group and \(H\) and \(K\) be two subgroups of \(G\). Then the normalizer of \(H\) in \(K\) is denoted by \(N_{K}(H) \) and defined by \(N_{K}(H)=\set{k\in K: kHk^{-1}=H} \).

Theorem-1

  Statement:
  Let \((G,\circ)\) be a group and \(H\) and \(K\) be two subgroups of \(G\). Then the normalizer \(N_{K}(H) \) of \(H\) in \(K\) is subgroup of \(K\).


  Proof:
  Given that \((G,\circ)\) is a group and \(H\) and \(K\) are two subgroups of \(G\).
  To prove \(N_{K}(H) \) is a subgroup of \(K\)

  • To prove \(N_{K}(H)\ne \phi \)
    Let \(e\) be the identity element of \(G\). Then \(e\in K\).
    Therefore \begin{align*} & eHe^{-1}=H \\ \implies & e\in N_{K}(H) \\ \implies & N_{K}(H)\ne \phi \end{align*}
  • To prove \(k_{1}\circ k_{2}\in N_{K}(H),~\forall~k_{1}, k_{2}\in N_{K}(H) \)
    Let \(k_{1}, k_{2}\in N_{K}(H) \) then \begin{align*} & \big(k_{1}\circ k_{2} \big)H\big(k_{1}\circ k_{2} \big)^{-1}\\ = & \big(k_{1}\circ k_{2} \big)H\big(k_{2}^{-1}\circ k_{1}^{-1} \big) \\ = & k_{1}\big( k_{2} Hk_{2}^{-1}\big) k_{1}^{-1} \\ = & k_{1}H k_{1}^{-1} ~\big[\because k_{2} Hk_{2}^{-1}=H \big] \\ = & H ~\big[\because k_{1} Hk_{1}^{-1}=H \big] \\ \end{align*} \(\therefore k_{1}\circ k_{2}\in N_{K}(H)\)
  • To prove \(k^{-1}\in N_{K}(H),~\forall~k\in N_{K}(H) \)
    Let \(k\in N_{K}(H) \) then \begin{align*} & \big(k^{-1}\big) H \big(k^{-1}\big)^{-1} \\ = & \big(k^{-1}\big) \big(k Hk^{-1}\big) k ~\big[\because k Hk^{-1}=H \big] \\ = & \big(k^{-1}\circ k\big) H\big(k^{-1}\circ k\big) \\ = & e He \\ = & H \end{align*} \(\therefore k^{-1}\in N_{K}(H)\)

  Hence \(N_{K}(H) \) is a subgroup of \(K\).

Theorem-2

  Statement:
  Let \((G,\circ)\) be a group and \(H\) and \(K\) be two subgroups of \(G\). Then the number of distinct conjugates of \(H\) induced by the elements of \(K\) is equal to \(\big[K:N_{K}(H) \big]\).


  Proof:
  Given that \((G,\circ)\) is a group and \(H\) and \(K\) are two subgroups of \(G\).
  Let \(T\) be the set of all distinct conjugates of \(H\) unduced by the elements of \(K\).
  Then we have \(T=\set{kHk^{-1}:k\in K}\).
  Let \(S\) be the set of all left coset of \(N_{K}(H) \) in \(K\).
  Then we have \(S=\set{kN_{K}(H):k\in K}\).
  Let us construct a mapping \(f:T\to S \) such that for \(k\in K\), \begin{align*} f(kHk^{-1})=kN_{K}(H) \end{align*}   To prove \(f\) is bijective

  • To prove \(f\) is well defined
    Let \(aHa^{-1},bHb^{-1}\in T\) where \(a,b\in K\) such that \begin{align*} & aHa^{-1}=bHb^{-1}\\ \implies & b^{-1}aHa^{-1}b^{-1}=H\\ \implies & \big(b^{-1}\circ a\big)H\big(b^{-1}\circ a\big)^{-1}=H\\ \implies & b^{-1}a\in N_{K}(H)\\ \implies & aN_{K}(H)=bN_{K}(H) \\ \implies & f(aHa^{-1})=f(bHb^{-1}) \end{align*} \(\therefore \) \(f\) is well defined.
  • To prove \(f\) is injective
    Let \(aHa^{-1},bHb^{-1}\in T\) where \(a,b\in K\) such that \begin{align*} & f(aHa^{-1})=f(bHb^{-1}) \\ \implies & aN_{K}(H)=bN_{K}(H) \\ \implies & b^{-1}a\in N_{K}(H)\\ \implies & \big(b^{-1}\circ a\big)H\big(b^{-1}\circ a\big)^{-1}=H\\ \implies & b^{-1}aHa^{-1}b^{-1}=H\\ \implies & aHa^{-1}=bHb^{-1} \end{align*} \(\therefore \) \(f\) is injective.
  • To prove \(f\) is surjective
    Let \(kN_{K}(H)\in S\) where \(k\in K\).
    Then \(kHk^{-1}\in T \) such that \(f(kHk^{-1})=kN_{K}(H) \). Therefore \(kHk^{-1} \) is a pre-image of \(kN_{K}(H)\) in \( T \). Since \(kN_{K}(H)\) is an arbitrary element of \(S\) ha a then each element of \(S\) has a pre-image in \( T \).
    \(\therefore \) \(f\) is surjective.

  Therefore \(f\) is bijective.
  Hence the number of distinct conjugates of \(H\) induced by the elements of \(K\) is equal to \(\big[K:N_{K}(H) \big]\).

Theorem-3

  Statement:
  Let \((G,\circ)\) be a group and \(H\) and \(K\) be two finite subgroups of \(G\). If \(H\) is invariant of under \(n\) elements of \(K\), then \(H\) has \(\frac{\big|K\big|}{n}\) conjugates by elements of \(K\).


  Proof:
  Given that \((G,\circ)\) is a group and \(H\) and \(K\) are two finite subgroups of \(G\).
  Let \(H\) is invariant of under \(n\) elements of \(K\).
  Then \(\big|N_{K}(H) \big|=n \).
  To prove \(H\) has \(\frac{\big|K\big|}{n}\) conjugates by elements of \(K\)
  We have that the the number of distinct conjugates of \(H\) induced by the elements of \(K\) is equal to \(\big[K:N_{K}(H) \big]\).
  Then to prove \(\big[K:N_{K}(H) \big]=\frac{\big|K\big|}{n}\)
  By Lagrange’s theorem \begin{align*} & \big|K\big|=\big[K:N_{K}(H) \big].\big|N_{K}(H)\big| \\ \implies & \big[K:N_{K}(H) \big]=\frac{\big|K\big|}{\big|N_{K}(H)\big|}\\ \implies & \big[K:N_{K}(H) \big]=\frac{\big|K\big|}{n}\\ \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

  Normalizer of a Subgroup plays a vital role in understanding group structure and its properties. Its relevance in Abstract Algebra ensures its importance in both academic and professional fields of Mathematics.

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 the Normalizer of a Subgroup?
    It is the set of elements in a group that commute with every element of a given subgroup under conjugation.
  2. Why is the Normalizer of a Subgroup important?
    It helps in understanding the structural properties of groups and their subgroups.
  3. Where is the concept of Normalizer of a Subgroup applied?
    It is applied in Abstract Algebra and other advanced mathematical studies.
  4. How is the Normalizer of a Subgroup defined?
    It is defined as the largest subgroup in which a given subgroup is normal.
  5. What are related topics to the Normalizer of a Subgroup?
    Topics include Mathematics and Abstract Algebra.
  6. Can the Normalizer of a Subgroup be equal to the group?
    Yes, when the subgroup is normal in the group.
  7. How is the Normalizer of a Subgroup calculated?
    It is determined by checking which elements commute under conjugation for all subgroup elements.
  8. Is the Normalizer of a Subgroup always a subgroup?
    Yes, it satisfies all subgroup properties.
  9. What is the relationship between Normalizers and Centralizers?
    The Centralizer is a subset of the Normalizer containing elements that commute with every element of the group.
  10. Where can I find practice questions on this topic?
    Practice questions can be found at Mathematics Questions and Abstract Algebra Questions.
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