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Chapter 14: Problem 3

Let \(H\) be a subgroup of \(G\), and assume that \(x H x^{-1}=H\) for all \(x \in G\). Then \(x^{-1} H x=H\) for all \(x \in G\), and \(H x=x H\) for all \(x \in G\).

Short Answer

Expert verified
Since \(x H x^{-1}=H\) for all \(x \in G\), we've proved that \(x^{-1} H x=H\) and \(H x = x H\) for all \(x \in G\). Thus, \(H\) is a normal subgroup of \(G\).

Step by step solution

01

Proving \(x^{-1} H x = H\)

Given that \(x H x^{-1} = H\) for all \(x \in G\), we want to show that \(x^{-1} H x=H\) for all \(x \in G\). Let \(y \in x^{-1} H x\). Then, \(y = x^{-1} h x\) for some \(h \in H\). Now, we can multiply on the left by \(x\) and on the right by \(x^{-1}\), so we have \[x y x^{-1} = x (x^{-1} h x) x^{-1}.\] Using the associative property of group operations, this can be rewritten as \[x y x^{-1} = (x x^{-1}) h (x x^{-1}) = h.\] Since \(x H x^{-1} = H\), we have \(h \in H\), and therefore \(x y x^{-1} \in H\). Thus, \(y \in x H x^{-1}\), which means that \(x^{-1} H x \subseteq H\). Now, let \(h' \in H\). Then, \(x^{-1} h' x \in x^{-1} H x\), and we already showed that \(x^{-1} H x \subseteq H\), meaning that \(x^{-1} h' x \in H\). Thus, \(H \subseteq x^{-1} H x\). Since both \(x^{-1} H x \subseteq H\) and \(H \subseteq x^{-1} H x\), we can conclude that \(x^{-1} H x = H\) for all \(x \in G\).
02

Proving \(H x = x H\)

Now we will show that \(H x = x H\) for all \(x \in G\). Let \(y \in H x\), so \(y = h x\) for some \(h \in H\). Since \(x H x^{-1} = H\), we have \[x^{-1} y = x^{-1} (h x) \in x^{-1} H x = H.\] Therefore, \(x^{-1} y \in H\) and can be written as \(x^{-1} y = h'\) for some \(h' \in H\). Multiplying on the left by \(x\), we have \[y = x h' \in x H.\] So, \(H x \subseteq x H\). Similarly, let \(z \in x H\), so \(z = x h''\) for some \(h'' \in H\). Since \(x^{-1} H x = H\), we have \[x^{-1} z = x^{-1} (x h'') \in x^{-1} H x = H.\] Therefore, \(x^{-1} z \in H\) and can be written as \(x^{-1} z = h'''\) for some \(h''' \in H\). Multiplying on the left by \(x\), we have \[z = x h''' \in x H.\] So, \(x H \subseteq H x\). Since both \(H x \subseteq x H\) and \(x H \subseteq H x\), we can conclude that \(H x = x H\) for all \(x \in G\). This implies that \(H\) is a normal subgroup of \(G\).

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Key Concepts

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

Group Theory
Group theory is a branch of mathematics that studies the algebraic structures known as groups. At its core, a group \(G\) consists of a set of elements along with an operation that combines any two elements to form another element of the set. This operation is often called 'group multiplication', even when the group elements aren't numbers. The importance of group theory arises from its ability to abstract and generalize the concept of symmetry in various mathematical contexts.

In the context of the aforementioned exercise, \(G\) represents a group, and we encounter a special type of subgroup within it, defined as a subset of elements that is also a group under the same operation. The exercise delves into the characteristics that distinguish an ordinary subgroup from a normal subgroup, which is significant for understanding more complex group structures and the way they behave.
Subgroups in Algebra
Subgroups are a fundamental concept in algebra and play a vital role in group theory. A subset \(H\) of a group \(G\) is called a subgroup if \(H\) itself is a group under the operation defined on \(G\). This means that subgroups must satisfy certain criteria, such as containing the group identity element, having the group operation be closed within \(H\), and every element in \(H\) must have an inverse within \(H\).

The exercise involves proving that the subgroup \(H\) is 'normal', which is a stronger condition. In algebra, a normal subgroup is one that is invariant under conjugation by elements of the group \(G\). This means that for every element \(x \in G\), the set \(x H x^{-1}\) is equal to \(H\), indicating a certain symmetry within the group structure that makes normal subgroups very interesting from an algebraic standpoint.
Associative Property
The associative property is a fundamental axiom in group theory, which states that the way in which elements are grouped during the group operation does not change the final result. In algebraic terms, for any three elements \(a, b,\) and \(c\) of a group \(G\), the property is expressed as \(a(bc) = (ab)c\). This property is crucial when performing manipulations within a group, as it allows for the rearrangement of parentheses in expressions, which is used repeatedly in the steps provided for the exercise.

When proving that \(H\) is a normal subgroup within the exercise, the associative property is used to show that elements remain within \(H\) when 'conjugated' by other elements of \(G\). The proofs for \(x^{-1} H x = H\) and \(H x = x H\) both crucially rely on the associative property, highlighting its importance in establishing the invariance of the subgroup \(H\), and thus its normality within the group \(G\).

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Most popular questions from this chapter

Let \(f: G \rightarrow G^{\prime}\) be a surjective homomorphism. Let \(H\) be a normal subgroup of \(G\). Show that \(f(H)\) is a normal subgroup of \(G^{\prime}\).

Let \(\omega_{1}, \ldots, \omega_{n}\) be roots of unity, and assume that if we let $$ A=\left(\begin{array}{cccc} \omega_{1} & 0 & \cdots & 0 \\ 0 & \omega_{2} & \cdots & 0 \\ \vdots & \ddots & \ddots & \vdots \\ 0 & \cdots & 0 & \omega_{n} \end{array}\right) $$ then \(A\) has period \(d .\) Let \(B\) be an invertible \(n \times n\) matrix. Show that \(B^{-1} A B\) also has period \(d\).

Let \(G\) be a group, and \(V\) s finite dimensional vector space over the field \(K\). By a representation of \(G\) on \(V\), one means a homomorphism \(\rho: G \rightarrow G L(V)\) of \(G\) into the group of invertible linear maps of \(V\) onto itself. The representation is said to be faithful if the kernel of \(\rho\) is the unit element of \(G\), i.e. if \(\rho\) is injective. If we select a basis of \(V\), then \(\rho\) amounts to a homomorphism of \(G\) into the group of matrices \(G L(n, K)\), where \(n=\operatorname{dim} V\). Show that a finite group \(G\) always admits such a representation, proceeding as follows. Let \(V\) be the vector space of formal linear combinations of elements of \(G\), as explained in the appendix of Chapter XIII. Thus the elements of \(G\) form a basis for this space, say \(\left\\{\sigma_{1}, \ldots, \sigma_{n}\right\\} .\) For each \(\sigma \in G\), let \(A\), be the linear map of \(V\) into itself such that $$ A_{\sigma}\left(\sigma_{i}\right)=\sigma \sigma_{i} $$ Show that the association \(\sigma \mapsto A\), is an injective homomorphism of \(G\) into \(G L(V)\)

Let \(G\) be a finite cyelic group of order \(n\). Let \(a\) be a generator. Let \(r\) be an integer \(\neq 0\), and relatively prime to \(n .\) Show that \(a_{r}\) is also a generator of G. Show that every generator of \(G\) ean be written in this form.

(a) Let \(V\) be a vector space over a field \(K\), and let \(v_{1}\) be an element of \(V\). Let \(G\) be the set of all invertible linear maps \(A\) of \(V\) into itself sueh that \(A v_{1}=v_{1}\). Show that \(G\) is a group. (b) More generally, let \(\left\\{v_{1}, \ldots, v_{i}\right\\}\) be a subset of \(V\). Let \(G\) be the set of all invertible operators \(A\) of \(V\) such that $$ A v_{1}=v_{1}, \ldots, A v_{i}=v_{i} $$ Show that \(G\) is a group. (c) Let \(S\) be a set, \(S^{\prime}\) s subset, and let \(G\) be the set of all maps \(f\) of \(S\) into itself such that \(f\) is invertible, and such that \(f(x)=x\) for all \(x \in S^{\prime}\). Show that \(G\) is a group.
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