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Chapter 9: Problem 42

Prove that \(\operatorname{Inn}(G)\) is a subgroup of \(\operatorname{Aut}(G)\).

Short Answer

Expert verified
To prove that the group of inner automorphisms, Inn(G), is a subgroup of the group of automorphisms, Aut(G), we must show three properties: 1. Closure under Composition: If we take two inner automorphisms, their composition is also an inner automorphism. This is shown by the composition of two inner automorphisms, 蠒_g and 蠒_h, resulting in another inner automorphism, (蠒_g 猕� 蠒_h)(x) = (gh)x(gh)^(-1). 2. Identity Element: The identity element of Aut(G) is also in Inn(G). This is proven by the identity function id_G being equal to the inner automorphism 蠒_e for the identity element e in G. 3. Inverses: Every element in Inn(G) has its inverse in Inn(G). This is shown by 蠒_g and 蠒_(g^(-1)) being inverses of each other, within the group Inn(G). Therefore, Inn(G) is a subgroup of Aut(G).

Step by step solution

01

Recall Definitions:

Recall that an automorphism of a group \(G\) is an isomorphism from \(G\) to itself, and the set of all automorphisms of \(G\) forms a group under the operation of composition of functions, denoted by \(\operatorname{Aut}(G)\). An inner automorphism is an automorphism of the form \(\varphi_g: G \rightarrow G\) defined by \(\varphi_g(x) = gxg^{-1}\) for some fixed \(g \in G\). The set of inner automorphisms of \(G\) is denoted by \(\operatorname{Inn}(G)\).
02

Closure under Composition:

Take two inner automorphisms \(\varphi_g, \varphi_h \in \operatorname{Inn}(G)\) for some \(g, h \in G\). We want to show that their composition, \(\varphi_g \circ \varphi_h\), is also an inner automorphism. For any \(x \in G\), observe that \[(\varphi_g \circ \varphi_h)(x) = \varphi_g(\varphi_h(x)) = \varphi_g(hxh^{-1}) = g(hxh^{-1})g^{-1} = (gh)x(gh)^{-1}.\] Thus, \(\varphi_g \circ \varphi_h\) is an inner automorphism corresponding to the element \(gh \in G\), so \(\operatorname{Inn}(G)\) is closed under composition.
03

Identity Element:

The identity element of \(\operatorname{Aut}(G)\) is the identity function, denoted as \(\operatorname{id}_G\). We need to show that \(\operatorname{id}_G\) is an element of \(\operatorname{Inn}(G)\). Observe that for the identity element \(e \in G\), the inner automorphism \(\varphi_e: G \rightarrow G\) is given by \(\varphi_e(x) = exe^{-1} = x\), for all \(x \in G\). Therefore, \(\operatorname{id}_G\) is an inner automorphism and belongs to \(\operatorname{Inn}(G)\).
04

Inverses in \(\operatorname{Inn}(G)\):

For any inner automorphism \(\varphi_g \in \operatorname{Inn}(G)\), we want to show that its inverse, \(\varphi_g^{-1}\), is also an inner automorphism. Observe that: \[(\varphi_g \circ \varphi_{g^{-1}})(x) = \varphi_g(\varphi_{g^{-1}}(x)) = \varphi_g(g^{-1}xg) = (gg^{-1})x(gg^{-1})^{-1} = x,\] and also \[(\varphi_{g^{-1}} \circ \varphi_g)(x) = \varphi_{g^{-1}}(gxg^{-1}) = (g^{-1}(gxg^{-1})g) = x,\] for all \(x \in G\). Thus, \(\varphi_g\) and \(\varphi_{g^{-1}}\) are inverses of each other, and \(\varphi_{g^{-1}} \in \operatorname{Inn}(G)\). This shows that every element in \(\operatorname{Inn}(G)\) has its inverse in \(\operatorname{Inn}(G)\). Since \(\operatorname{Inn}(G)\) is closed under composition, contains the identity element, and every element has its inverse in \(\operatorname{Inn}(G)\), we conclude that \(\operatorname{Inn}(G)\) is a subgroup of \(\operatorname{Aut}(G)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inner Automorphisms
An inner automorphism is a type of transformation within a group that reflects the idea of twisting or rotating the group elements in a particular way. Imagine you have a group \(G\), a mathematical set equipped with an operation that fulfills specific properties like closure and associativity, among others. If you pick an element \(g\) from this group \(G\), you can define an inner automorphism \(\varphi_g\). This is a function that will take any element \(x\) in the group and transform it via the formula \(\varphi_g(x) = gxg^{-1}\). Here鈥檚 what this means:
  • \(gx\) multiplies \(x\) by \(g\) from the left.
  • Then, you multiply the result by \(g^{-1}\) from the right, effectively twisting \(x\) around \(g\).
In essence, inner automorphisms use the group's elements to shuffle or permute itself. A crucial point to understand is that these transformations are special because they arise directly from the elements of the group itself rather than any external or arbitrary rules.
Group Theory
Group theory is a fundamental area of study in abstract algebra that investigates the algebraic structures known as groups. A group is simply a set of elements combined with an operation that satisfies four essential properties: closure, associativity, identity, and inverses.
  • Closure: Whenever you combine two elements from the group using the operation, the result is still an element of the group.
  • Associativity: The way you group the elements during the operation does not affect the outcome, meaning \((a \cdot b) \cdot c = a \cdot (b \cdot c)\).
  • Identity Element: There exists a unique element in the group such that when it is combined with any element in the group, the element remains unchanged.
  • Inverse Element: For every element, there exists another element in the group so that when combined, the result is the identity element.
Understanding these properties helps build the foundation for more complex concepts, such as symmetries in mathematics and the solutions to polynomial equations. Group theory is used across multiple fields of science and technology, aiding in solving modern problems.
Automorphism Group
An automorphism is a structure-preserving map from a mathematical object back onto itself, much like a self-mirror where the essence of the object remains intact. When we specifically talk about groups, an automorphism is an isomorphism from a group \(G\) to itself, meaning it is a bijective map preserving the group operation.
All such automorphisms of a group \(G\) together form its automorphism group, denoted by \(\operatorname{Aut}(G)\). Here鈥檚 why this is significant:
  • \(\operatorname{Aut}(G)\) itself is a group.
  • This means it satisfies the group properties鈥攃losure, associativity, contains an identity, and has inverses for every element.
  • The automorphism group allows mathematicians to explore the symmetries and the internal structure of \(G\) in great depth.
In particular, the inner automorphisms, discussed earlier, form a subgroup of \(\operatorname{Aut}(G)\) called \(\operatorname{Inn}(G)\). The relationship and interactions between \(\operatorname{Inn}(G)\) and \(\operatorname{Aut}(G)\) offer rich insights into the nature and behavior of the group \(G\).

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Most popular questions from this chapter

Let \(G=\mathbb{R} \backslash\\{-1\\}\) and define a binary operation on \(G\) by $$a * b=a+b+a b$$ Prove that \(G\) is a group under this operation. Show that \((G, *)\) is isomorphic to the multiplicative group of nonzero real numbers.

Prove that \(U(8)\) is isomorphic to the group of matrices $$\left(\begin{array}{ll}1 & 0 \\\0 & 1\end{array}\right),\left(\begin{array}{cc}1 & 0 \\\0 & -1 \end{array}\right),\left(\begin{array}{cc}-1 & 0 \\\0 & 1 \end{array}\right),\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)$$

Groups of order \(2 p .\) In this series of exercises we will classify all groups of order \(2 p,\) where \(p\) is an odd prime. (a) Assume \(G\) is a group of order \(2 p,\) where \(p\) is an odd prime. If \(a \in G,\) show that \(a\) must have order \(1,2, p,\) or \(2 p\). (b) Suppose that \(G\) has an element of order \(2 p\). Prove that \(G\) is isomorphic to \(\mathbb{Z}_{2 p}\). Hence, \(G\) is cyclic. (c) Suppose that \(G\) does not contain an element of order \(2 p .\) Show that \(G\) must contain an element of order \(p .\) Hint: Assume that \(G\) does not contain an element of order \(p\). (d) Suppose that \(G\) does not contain an element of order \(2 p .\) Show that \(G\) must contain an element of order 2 .(e) Let \(P\) be a subgroup of \(G\) with order \(p\) and \(y \in G\) have order 2 . Show that \(y P=P y\). (f) Suppose that \(G\) does not contain an element of order \(2 p\) and \(P=\langle z\rangle\) is a subgroup of order \(p\) generated by \(z\). If \(y\) is an element of order 2 , then \(y z=z^{k} y\) for some \(2 \leq k

An automorphism of a group \(G\) is an isomorphism with itself. Prove that complex conjugation is an automorphism of the additive group of complex numbers; that is, show that the map \(\phi(a+b i)=a-b i\) is an isomorphism from \(\mathbb{C}\) to \(\mathbb{C}\).

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