Inner Automorphism in Group Theory: Definition and Key Theorems
Inner Automorphism in Group Theory
Inner Automorphisms in group theory play a crucial role in Group Theory, one of the most fundamental branches of abstract algebra. This article delves into the definition of inner automorphisms and explores key theorems associated with them. It highlights their significance in courses like B.Sc. and M.Sc. Mathematics and competitive exams such as JEE, GATE, and GRE. Additionally, we discuss real-life applications and how this topic helps students in advanced studies and research.
What You Will Learn?
In this post, you will explore::
- Definition: Inner Automorphism
- Theorems:
- Let \(\pmb{(G,\cdot)}\) be a group. Then \(\pmb{Inn(G)}\) is a subgroup of \(\pmb{Aut(G)}\).
- Let \(\pmb{(G,\cdot)}\) be a group. Then \(\pmb{Inn(G)}\) is normal subgroup of \(\pmb{Aut(G)}\).
- Let \(\pmb{(G,\cdot)}\) be a group and \(\pmb{Z(G)}\) be the center of \(\pmb{G}\). Then \(\pmb{\frac{G}{Z(G)}\backsimeq Inn(G)}\).
- Let \(\pmb{(G,\cdot)}\) be a group and \(\pmb{H}\) be a subgroup of \(\pmb{G}\). Let \(\pmb{N(H)}\) and \(\pmb{C(H)}\) be the normalizer and centralizer of \(\pmb{H}\). Then \(\pmb{\frac{N(H)}{C(H)}\backsimeq}\) a subgroup of \(\pmb{Aut(G)}\).
Things to Remember
Before diving into this post, make sure you are familiar with:
Basic Definitions and Concepts of
- Set Theory
- Relations
- Mappings
- Group Theory
Introduction
Inner Automorphism
Definition:
Let \(\pmb{(G,\cdot)}\) be group. The set of all automorphisms \(\pmb{\theta_{a}:G\to G~\forall~ a\in G}\), defined by \(\pmb{\theta_{a}(x)=a\cdot x\cdot a^{-1}~\forall~x\in G}\), is said to be the Inner Automorphism of \(\pmb{G}\) and is denoted by \(\pmb{Inn(G)}\).
i.e.,\(\pmb{Inn(G)=\{\theta_{a}~|~ \theta_{a}:G \to G~\text{defined by}~\theta_{a}(x)=a\cdot x\cdot a^{-1}~\forall~x\in G~\text{and}~a\in G \} }\)
Theorem-1
Statement:
Let \(\pmb{(G,\cdot)}\) be a group. Then \(\pmb{Inn(G)}\) is a subgroup of \(\pmb{Aut(G)}\).
Proof:
Given that \(\pmb{(G,\cdot)}\) is a group.
To prove \(\pmb{Inn(G)}\) is a subgroup of \(\pmb{Aut(G)}\).
We have \(\pmb{Inn(G)=\{\theta_{a}~|~a\in G \} }\) where \(\pmb{\theta_{a}:G\to G}\) defined by \(\pmb{\theta_{a}(x)=a\cdot x\cdot a^{-1}~\forall~x\in G}\).
- Let \( \pmb{\theta_{a}\in Inn(G)} \)
But \( \pmb{\theta_{a}} \) is an automorphism then \(\pmb{\theta_{a}\in Aut(G) }\)
Therefore \( \pmb{Inn(G)\subseteq Aut(G)} \) - To prove \(\pmb{Inn(G)\ne \Phi}\)
Let \(\pmb{e}\) is identity member of \(\pmb{G}\) and \(\pmb{I_{G}}\) is the identity mapping on \(\pmb{G}\). Then \(\pmb{\forall~x\in G}\) \begin{align*} \pmb{\theta_{e}(x)}=& \pmb{e\cdot x\cdot e^{-1}}\\ =& \pmb{x}\\ =& \pmb{I_{G}(x)} \end{align*} So \(\pmb{\theta_{e}=I_{G} \implies I_{G}\in Inn(G)}\).
Therefore \(\pmb{Inn(G)\ne \Phi}\). - To prove \(\pmb{\theta_{a}\circ (\theta_{b})^{-1}\in Inn(G)~\forall~\theta_{a},\theta_{b}\in Inn(G)}\)
Let \(\pmb{\theta_{a},\theta_{b}\in Inn(G) }\) where \(\pmb{a,b\in G}\) then \(\pmb{\theta_{a}(x)}=\pmb{a\cdot x\cdot a^{-1} } \) and \(\pmb{\theta_{b}(x)}=\pmb{b\cdot x\cdot b^{-1} } \) \(\pmb{\forall~x\in G}\) \begin{align*} \pmb{(\theta_{a}\circ (\theta_{b})^{-1})(x) }=&\pmb{(\theta_{a}\circ \theta_{b^{-1}})(x) }\\ =&\pmb{\theta_{a\cdot b^{-1} }(x) } \end{align*} Therefore \(\pmb{\theta_{a}\circ (\theta_{b})^{-1}= \theta_{a\cdot b^{-1} } \in Inn(G)}\) Since \(\pmb{a\cdot b^{-1}\in G }\)
Hence \(\pmb{Inn(G)}\) is a subgroup of \(\pmb{Aut(G)}\).
Theorem-2
Statement:
Let \(\pmb{(G,\cdot)}\) be a group. Then \(\pmb{Inn(G)}\) is normal subgroup of \(\pmb{Aut(G)}\).
Proof:
Given that \(\pmb{(G,\cdot)}\) is a group.
To prove \(\pmb{Inn(G)}\) is normal in \(\pmb{Aut(G)}\).
Let \( \pmb{\theta_{a}\in Inn(G)} \) where \(\pmb{a\in G}\) and Let \( \pmb{\alpha\in Aut(G)} \). Then \(\pmb{\forall~x\in G}\)
\begin{align*}
\pmb{(\alpha\circ \theta_{a}\circ \alpha^{-1})(x)}=& \pmb{\alpha( \theta_{a}( \alpha^{-1}(x))}\\
=& \pmb{\alpha(a\cdot \alpha^{-1}(x)\cdot a^{-1} )}\\
=& \pmb{\alpha(a)\cdot \alpha(\alpha^{-1}(x))\cdot \alpha(a^{-1})}\\
=& \pmb{\alpha(a)\cdot (\alpha\circ\alpha^{-1})(x)\cdot (\alpha(a))^{-1}}\\
=& \pmb{\alpha(a)\cdot I_{G}(x)\cdot (\alpha(a))^{-1}}\\
=& \pmb{\alpha(a)\cdot x\cdot (\alpha(a))^{-1}}\\
=& \pmb{\theta_{\alpha(a)}(x)}\\
\end{align*}
So \(\pmb{\alpha\circ \theta_{a}\circ \alpha^{-1}= \theta_{\alpha(a)} }\).
Since \(\pmb{\alpha(a)\in G}\) then \(\pmb{\theta_{\alpha(a)}\in Inn(G) }\)
Therefore \(\pmb{\alpha\circ \theta_{a}\circ \alpha^{-1}\in Inn(G) }\)
Hence \(\pmb{Inn(G)}\) is normal in \(\pmb{Aut(G)}\).
Theorem-3
Statement:
Let \(\pmb{(G,\cdot)}\) be a group and \(\pmb{Z(G)}\) be the center of \(\pmb{G}\). Then \(\pmb{\frac{G}{Z(G)}\backsimeq Inn(G)}\).
Proof:
Given that \(\pmb{(G,\cdot)}\) is a group and \(\pmb{Z(G)}\) is the center of \(\pmb{G}\).
Then we have \(\pmb{Z(G)=\{x\in G~|~x\cdot g=g\cdot x~\forall~g\in G \}}\).
Also, \(\pmb{Z(G)}\) is normal in \(\pmb{G}\) therefore \(\pmb{\frac{G}{Z(G)}}\) exists.
To prove \(\pmb{\frac{G}{Z(G)}\backsimeq Inn(G)}\).
Let us construct a mapping \(\pmb{\phi:G\to Inn(G)}\) such that \(\pmb{\phi(a)=\theta_{a}~\forall~a\in G}\) where \(\pmb{\theta_{a}:G\to G}\) is defined by \(\pmb{\theta_{a}(x)=a\cdot x \cdot a^{-1}~\forall~x\in G}\).
- To prove \(\pmb{\phi}\) is well defined.
Let \(\pmb{a,b\in G }\) such that \begin{align*} \pmb{a=b\implies \theta_{a}=\theta_{b}\implies \phi(a)=\phi(b) } \end{align*} Therefore \(\pmb{\phi}\) is well defined. - To prove \(\pmb{\phi}\) is surjective.
Let \(\pmb{\theta_{a}\in Inn(G) }\) where \(\pmb{a\in G}\).
Then \(\pmb{\phi(a)=\theta_{a}}\).
Therefore \(\pmb{a}\) is a pre-image of \(\pmb{\theta_{a}}\). Since \(\pmb{\theta_{a}}\) is an arbitrary member of \(\pmb{Inn(G)}\), therefore each member of \(\pmb{Inn(G)}\) has pre-image in \(\pmb{G}\).
Hence \(\pmb{\phi}\) is surjective. - To prove \(\pmb{\phi}\) is homomorphism.
Let \(\pmb{a,b\in G }\). Then \begin{align*} \pmb{\phi(a\cdot b)= \theta_{a\cdot b}= \theta_{a}\circ \theta_{b}=\phi(a)\circ\phi(b)} \end{align*} Hence \(\pmb{\phi}\) is homomorphism. - To prove \(\pmb{Ker~\phi=z(G)}\).
- To prove \(\pmb{Ker~\phi \subseteq z(G)}\).
Let \(\pmb{a\in Ker~\phi }\) and \(\pmb{g\in G }\). Now
\begin{align*} &\pmb{a\in Ker~\phi}\\ \implies& \pmb{\phi(a)=I_{G}}\\ \implies& \pmb{\theta_{a}=I_{G}}\\ \implies& \pmb{\theta_{a}(g)=I_{G}(g)}\\ \implies&\pmb{a\cdot g \cdot a^{-1}=g}\\ \implies&\pmb{a\cdot g=g\cdot a}\\ \implies&\pmb{a\in z(G)} \end{align*} Therefore \(\pmb{Ker~\phi \subseteq z(G)}\). - To prove \(\pmb{z(G) \subseteq Ker~\phi}\).
Let \(\pmb{b\in z(G) }\) and \(\pmb{g\in G }\). Now \begin{align*} &\pmb{a\in z(G)}\\ \implies&\pmb{a\cdot g=g\cdot a}\\ \implies&\pmb{a\cdot g \cdot a^{-1}=g}\\ \implies& \pmb{\theta_{a}(g)=I_{G}(g)}\\ \implies& \pmb{\theta_{a}=I_{G}}\\ \implies& \pmb{\phi(a)=I_{G}}\\ \implies&\pmb{a\in Ker~\phi}\\ \end{align*} Therefore \(\pmb{z(G) \subseteq Ker~\phi}\).
- To prove \(\pmb{Ker~\phi \subseteq z(G)}\).
Therefore by the isomorphism theorem, \(\pmb{\frac{G}{Z(G)}\backsimeq Inn(G)}\).
Theorem-4
Statement:
Let \(\pmb{(G,\cdot)}\) be a group and \(\pmb{H}\) be a subgroup of \(\pmb{G}\). Let \(\pmb{N(H)}\) and \(\pmb{C(H)}\) be the normalizer and centralizer of \(\pmb{H}\). Then \(\pmb{\frac{N(H)}{C(H)}\backsimeq}\) a subgroup of \(\pmb{Aut(G)}\).
Proof:
Given that \(\pmb{(G,\cdot)}\) is a group and \(\pmb{H}\) is a subgroup of \(\pmb{G}\). Also \(\pmb{N(H)}\) and \(\pmb{C(H)}\) are normalizer and centralizer of \(\pmb{H}\).
Then we have \(\pmb{N(H)=\{x\in G~|~x\cdot H\cdot x^{-1}=H \}}\) and \(\pmb{C(H)=\{x\in G~|~x\cdot h\cdot x^{-1}=h~\forall~h\in H \}}\).
To prove \(\pmb{\frac{N(H)}{C(H)}\backsimeq}\) a subgroup of \(\pmb{Aut(G)}\).
Let us construct a mapping \(\pmb{f:N(H)\to Aut(H)}\) such that \(\pmb{f(a)=\theta_{a}~\forall~a\in N(H)}\) where \(\pmb{\theta_{a}|_{H}:H\to H}\) defined by \(\pmb{\theta_{a}|_{H}(x)=a\cdot x \cdot a^{-1}~\forall~x\in H}\). And we know that \(\pmb{Aut(H)}\) is a subgroup of \(\pmb{Aut(G)}\). Now
- To prove \(\pmb{f}\) is well defined.
Let \(\pmb{a,b\in N(H)}\). Such that \(\pmb{a=b}\). \begin{align*} &\pmb{a=b}\\ \implies&\pmb{a\cdot x \cdot a^{-1}=b\cdot x \cdot b^{-1}~\forall~x\in H}\\ \implies&\pmb{\theta_{a}|_{H}(x)=\theta_{b}|_{H}(x)~\forall~x\in H}\\ \implies&\pmb{\theta_{a}|_{H}=\theta_{b}|_{H}}\\ \implies&\pmb{f(a)=f(b)}\\ \end{align*} Therefore \(\pmb{f}\) is well defined. - To prove \(\pmb{f}\) is surjective.
Let \(\pmb{\theta_{h}|_{H}\in Aut(H)}\) where \(\pmb{h\in H}\). Then \(\pmb{h\in N(H)}\) since \(\pmb{H\subseteq N(H)}\). So that \(\pmb{f(h)= \theta_{h}|_{H}}\).
Implies \(\pmb{h}\) is a pre-image of \(\pmb{\theta_{h}|_{H}}\) in \(\pmb{N(H)}\). Since \(\pmb{\theta_{h}|_{H}}\) is an arbitrary member of \(\pmb{Aut(H)}\), therefore each member of \(\pmb{Aut(H)}\) has a pre-image in \(\pmb{N(H)}\).
Hence \(\pmb{f}\) is surjective. - To prove \(\pmb{f}\) is homomorphism.
Let \(\pmb{a,b\in N(H)}\). Then \begin{align*} \pmb{f(a\cdot b)}&=\pmb{\theta_{a\cdot b}|_{H}}\\ &=\pmb{\theta_{a}|_{H}\circ \theta_{b}|_{H}}\\ &=\pmb{f(a)\circ f(b)}\\ \end{align*} Therefore \(\pmb{f}\) is homomorphism. - To prove \(\pmb{Ker~f=C(H)}\)
Let \(\pmb{I_{H}}\) be the identity mapping on \(\pmb{H}\)- To prove \(\pmb{Ker~f\subseteq C(H)}\)
Let \begin{align*} &\pmb{a\in Ker~f}\\ \implies&\pmb{f(a)=I_{H}}\\ \implies&\pmb{\theta_{a}|_{H}=I_{H}}\\ \implies&\pmb{\theta_{a}|_{H}(x)=I_{H}(x)~\forall~ x\in H}\\ \implies&\pmb{a\cdot x \cdot a^{-1}=x~\forall~ x\in H}\\ \implies&\pmb{a\in C(H)}\\ \end{align*} Therefore \(\pmb{Ker~f\subseteq C(H)}\). - To prove \(\pmb{C(H)\subseteq Ker~f}\)
Let \begin{align*} &\pmb{c\in C(H)}\\ \implies&\pmb{c\cdot x \cdot c^{-1}=x~\forall~ x\in H}\\ \implies&\pmb{\theta_{c}|_{H}(x)=I_{H}(x)~\forall~ x\in H}\\ \implies&\pmb{\theta_{c}|_{H}=I_{H}}\\ \implies&\pmb{c\in Ker~f} \end{align*} Threfore \(\pmb{C(H)\subseteq Ker~f}\).
- To prove \(\pmb{Ker~f\subseteq C(H)}\)
Then by the isomorphism theorem, \(\pmb{\frac{N(H)}{C(H)}\backsimeq}\) a subgroup of \(\pmb{Aut(G)}\).
Applications
Understanding inner automorphisms has applications beyond group theory. In physics, they help explain symmetries in physical systems, while in economics, group theory models complex systems and interactions. They also find applications in numerical methods, where transformations preserve certain structures and properties.
Conclusion
In conclusion, inner automorphisms offer an insightful understanding of group structures and their automorphisms. The theorems discussed in this article provide a strong foundation for further study and research in Abstract Algebra and its applications. Students preparing for exams or pursuing research in mathematics must grasp these concepts thoroughly.
References
- Algebra by Michael Artin
- Abstract Algebra by David S. Dummit and Richard M. Foote
- Research Paper on Automorphism Groups
- A Course in Group Theory by John F. Humphreys
- Algebraic Structures by Richard G. Swan
- Mathematics Stack Exchange on Automorphisms
- Journal of Algebra on Inner Automorphisms
- Group Theory by Joseph J. Rotman
- Isomorphisms and Groups – Springer Link
- Automorphisms and Group Actions by László Lovász
Related Articles
- Normal Subgroups and their Properties
- Homomorphism
- Automorphism
- Introduction to Abstract Algebra
FAQs
- What is an inner automorphism? An inner automorphism is a transformation of a group element by conjugation within the group.
- What is the significance of inner automorphisms in group theory? Inner automorphisms help in understanding the structure of groups and their symmetry properties.
- How do inner automorphisms relate to the center of a group? The set of all inner automorphisms is isomorphic to the quotient of the group by its center.
- What is the difference between inner and outer automorphisms? Inner automorphisms are generated by elements within the group, while outer automorphisms involve transformations not expressible as inner automorphisms.
- What is the center of a group? The center of a group consists of all elements that commute with every other element of the group.
- How are inner automorphisms used in physics? In physics, they describe symmetries in quantum mechanics and gauge theory.
- What is the normalizer of a subgroup? The normalizer of a subgroup consists of all elements in the group that stabilize the subgroup under conjugation.
- How do centralizers differ from normalizers? Centralizers consist of all elements that commute with every element of the subgroup, while normalizers stabilize the subgroup.
- What are automorphism groups? Automorphism groups consist of all bijective mappings from a group to itself that preserve the group operation.
- How can I study inner automorphisms for exams like JEE and GATE? Understanding their definitions, key theorems, and applications will help you tackle questions on these topics in competitive exams.
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