Conjugate of a Subgroup

Conjugate of a Subgroup

The Conjugate of a Subgroup is an essential concept in Mathematics and Abstract Algebra. It plays a critical role in Group Theory, providing insights into subgroup relationships within a group. Historically, this concept has been studied to understand the symmetries of algebraic structures and has applications in modern mathematical research.

What You Will Learn?

  In this post, you will explore:

  • Definition: Conjugate of a Subgroup
  • Theorem-1: Let \((G,\circ)\) be a group and \(H\) be a subgroup of \(G\) and \(a\in G\). Then the conjugate \(aHa^{-1}\) of \(H\) is a subgroup of \(G\).
  • Theorem-2: Let \((G,\circ)\) be a group and \(H\) be a subgroup of \(G\) and \(a\in G\). Then prove that \(H \cong aHa^{-1} \).

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

  The Conjugate of a Subgroup is defined as a subgroup transformed by an element of the parent group. This concept is pivotal in Abstract Algebra, helping classify subgroups and explore their properties. In Abstract Algebra Questions, it is often utilized to solve problems involving group structures and symmetries.

Conjugate of a Subgroup

  Definition:
Let \((G,\circ)\) be a group and \(H\) be a subgroup of \(G\) and \(a\in G\). Then the set \(aHa^{-1}=\set{a\circ h \circ a^{-1}: h\in H}\) is said to be a conjugate of \(H\).

Theorem-1

  Statement:
  Let \((G,\circ)\) be a group and \(H\) be a subgroup of \(G\) and \(a\in G\). Then the conjugate \(aHa^{-1}\) of \(H\) is a subgroup of \(G\).
  In addition \(H \cong aHa^{-1} \)


  Proof:
  Given that \((G,\circ)\) is a group and \(H\) is a subgroup of \(G\) and \(a\in G\).
  To prove \(aHa^{-1}\) is a subgroup of \(G\)

  • To prove \(aHa^{-1}\ne \phi\)
    Let \(e\) is the identity element of \(G\). Then \(e\in H\).
    Therefore \begin{align*} & a\circ e \circ a^{-1}\in aHa^{-1}\\ \implies & a\circ a^{-1}\in aHa^{-1} \\ \implies & e\in aHa^{-1} \\ \end{align*} \(\therefore aHa^{-1}\ne \phi\).
  • To prove \(x\circ y \in aHa^{-1}~\forall~x,y \in aHa^{-1}\)
    Let \(x,y \in aHa^{-1} \) then there exists \(h_{1},h_{2}\in H \) such that \(x=a\circ h_{1} \circ a^{-1} \) and \(y=a\circ h_{2} \circ a^{-1} \) \begin{align*} & x\circ y= \big( a\circ h_{1} \circ a^{-1} \big)\circ \big(a\circ h_{2} \circ a^{-1} \big)\\ \implies & x\circ y= a \circ h_{1} \circ \big(a^{-1} \circ a \big )\circ h_{2} \circ a^{-1} \\ \implies & x\circ y= a \circ \big( h_{1} \circ h_{2} \big)\circ a^{-1} \\ \implies & x\circ y \in aHa^{-1}~\big[\because h_{1} \circ h_{2}\in H \big] \end{align*}
  • To prove \(x^{-1} \in aHa^{-1}~\forall~x\in aHa^{-1}\)
    Let \(x \in aHa^{-1} \) then there exists \(h\in H \) such that \begin{align*} & x=a\circ h \circ a^{-1} \\ \implies & x^{-1}=\big( a\circ h \circ a^{-1}\big)^{-1} \\ \implies & x^{-1}=\big( a^{-1}\big)^{-1} \circ h^{-1}\circ a^{-1} \\ \implies & x^{-1}=a \circ h^{-1}\circ a^{-1} \\ \implies & x^{-1}\in aHa^{-1}~\big[\because h^{-1} \in H \big] \end{align*}

  Hence \(aHa^{-1}\) is a subgroup of \(G\).

Theorem-2

  Statement:
  Let \((G,\circ)\) be a group and \(H\) be a subgroup of \(G\) and \(a\in G\). Then prove that \(H \cong aHa^{-1} \).


  Proof:
  Given that \((G,\circ)\) is a group and \(H\) is a subgroup of \(G\) and \(a\in G\).
  To prove \(H \cong aHa^{-1} \)
  Let us construct a mapping \(f:H\to aHa^{-1} \) such that for \(h\in H\) \begin{align*} f(h)=a\circ h \circ a^{-1} \end{align*}

  • To prove \(f\) is well defined
    Let \(h_{1},h_{2}\in H \) such that \begin{align*} & h_{1}=h_{2} \\ \implies & a\circ h_{1} \circ a^{-1}=a\circ h_{2} \circ a^{-1} \\ \implies & f(h_{1})=f(h_{2}) \end{align*} \(\therefore\) \(f\) is well defined.
  • To prove \(f\) is injective
    Let \(h_{1},h_{2}\in H \) such that \begin{align*} & f(h_{1})=f(h_{2})\\ \implies & a\circ h_{1} \circ a^{-1}=a\circ h_{2} \circ a^{-1} \\ \implies & h_{1}=h_{2} \end{align*} \(\therefore\) \(f\) is injective.
  • To prove \(f\) is surjective
    Let \(a\circ h_{3} \circ a^{-1}\in aHa^{-1} \) for some \(h_{3}\in H \).
    Then \(f(h_{3})=a\circ h_{3} \circ a^{-1}\).
    \(\therefore\) \(h_{3} \) is a pre-image of \(a\circ h_{3} \circ a^{-1} \). Since \(h_{3} \) is an arbitrary element of \(aHa^{-1}\), then each element of \(aHa^{-1}\) has a pre-image in \(H\).
    \(\therefore\) \(f\) is surjective.
  • To prove \(f\) is homomorphism
    Let \(h_{4},h_{5}\in H \) then \begin{align*} f\big( h_{4}\circ h_{5}\big) & = a\circ \big( h_{4}\circ h_{5}\big) \circ a^{-1}\\ & = \big( a\circ h_{4} \big) \circ \big( a^{-1} \circ a \big) \circ \big(h_{5} \circ a^{-1}\big) \\ & = \big( a\circ h_{4} \circ a^{-1}\big) \circ \big( a \circ h_{5} \circ a^{-1}\big) \\ & = f\big( h_{4} \big) \circ f\big( h_{5} \big) \end{align*} \(\therefore\) \(f\) is homomorphism.

  \( \therefore f\) is an isomorphism.
  Hence \(H \cong aHa^{-1} \).

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 study of the Conjugate of a Subgroup reveals significant structural information about groups. This concept underpins numerous results in Mathematics and is indispensable in advanced studies of Group Theory. Its applications continue to enrich the understanding of algebraic systems in academic and research settings.

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 conjugate of a subgroup?
    It is a subgroup formed by applying a conjugation operation with an element of the parent group.
  2. How is the conjugate of a subgroup defined?
    The conjugate of a subgroup \( H \) by an element \( g \) is \( gHg^{-1} \), where \( g \) belongs to the group.
  3. Why are conjugate subgroups important?
    They provide insights into the symmetry and structure of groups.
  4. What is the relationship between normal subgroups and conjugates?
    A subgroup is normal if its conjugate by any group element equals itself.
  5. Where is the concept of conjugate subgroups applied?
    It is applied in studying group actions, normality, and group homomorphisms.
  6. What is an example of a conjugate subgroup?
    For \( H = \{e, a\} \) in \( G = \{e, a, b, ab\} \), the conjugate by \( b \) is \( bHb^{-1} \).
  7. Can conjugate subgroups be equal?
    Yes, especially if the subgroup is normal in the parent group.
  8. How can I practice problems on conjugate subgroups?
    Problems are available in Mathematics Questions.
  9. What is the role of conjugate subgroups in group homomorphisms?
    Conjugates help understand kernel properties and quotient groups.
  10. Where can I find detailed notes on this topic?
    Detailed notes are available in Group Theory Notes.
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