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Chapter 3: Problem 2

Zeigen Sie: In einer Gruppe sind die Gleichungen \(x * a=b\) und \(a * y=b\) eindeutig nach \(x\) bzw. \(y\) aufl枚sbar.

Short Answer

Expert verified
Question: In a group, show that the equations \(x * a = b\) and \(a * y = b\) are uniquely solvable for \(x\) and \(y\), respectively. Answer: In a group, the equations \(x * a = b\) and \(a * y = b\) are uniquely solvable for \(x\) and \(y\), respectively, with the solutions \(x = b * a^{-1}\) and \(y = a^{-1} * b\).

Step by step solution

01

Find the relationships between x and y

Since we are given that \(x * a = b\) and \(a * y = b\), we can use these equations to determine the relationships between \(x\) and \(y\). From the first equation, we have: $$x * a = b$$ From the second equation, we have: $$a * y = b$$
02

Apply the inverse property

A group must have an inverse for each element. Since \(a\) is an element of the group, we know there must be an element \(a^{-1}\) such that: $$ a * a^{-1} = a^{-1} * a = e $$ Where \(e\) is the identity element. Now, let's multiply both sides of our first equation by \(a^{-1}\) on the right side: $$ (x * a) * a^{-1} = b * a^{-1} $$ Since the group is associative, we can rearrange this as: $$ x * (a * a^{-1}) = b * a^{-1} $$ Now, using the inverse property, we can further simplify this to: $$ x * e = b * a^{-1} $$ Since \(e\) is the identity element, \(x * e = x\). So, we have: $$ x = b * a^{-1} $$
03

Solve for y

Similarly, multiply both sides of the second equation by \(a^{-1}\) on the left side. $$ a^{-1} * (a * y) = a^{-1} * b $$ Using associative property, we can rearrange this as: $$ (a^{-1} * a) * y = a^{-1} * b $$ And using the inverse property: $$ e * y = a^{-1} * b $$ Since \(e\) is the identity element, \(e * y = y\). So, we have: $$ y = a^{-1} * b $$
04

Conclusion

Based on the above steps, we have found unique solutions for \(x\) and \(y\): $$ x = b * a^{-1} \text{ and } y = a^{-1} * b $$ We have shown that in a group, the equations \(x * a = b\) and \(a * y = b\) are uniquely solvable for \(x\) and \(y\), respectively.

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Key Concepts

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

Inverse Property
The inverse property is a foundational concept in group theory. It ensures that for any element in a group, there exists another element that can "undo" its operation. This unique inverse element is crucial for solving equations within a group.

Here's why the inverse is so important:
  • Every element in a group must have an inverse. If you have an element \(a\), there must be another element \(a^{-1}\) such that when \(a\) is combined with \(a^{-1}\), the result is the identity element \(e\).
  • The relationship is expressed mathematically as: \(a * a^{-1} = a^{-1} * a = e\), where \(e\) is the identity element.
  • This property allows us to "cancel out" elements. For example, to solve \(x * a = b\) for \(x\), multiply both sides by \(a^{-1}\), yielding \(x = b * a^{-1}\).
Understanding the inverse property is key to manipulating equations in groups effectively.
Associative Property
The associative property is another essential facet of group theory that simplifies working with group elements. This property guarantees the predictable behavior of operations within a group.

Let's delve into why it's significant:
  • The associative property states that the grouping of operations does not affect their result. Mathematically, for any elements \(a\), \(b\), and \(c\), the equation \((a * b) * c = a * (b * c)\) holds true.
  • This property is crucial for simplifying expressions. For instance, when solving \((x * a) * a^{-1} = b * a^{-1}\), we can rewrite it as \(x * (a * a^{-1}) = b * a^{-1}\) due to associativity.
  • Without associativity, calculations in groups would become cumbersome and unpredictable, making it difficult to solve equations or study group structure.
With the associative property, you can confidently rearrange and simplify group operations.
Identity Element
The identity element is a pillar of group theory, providing a neutral starting point for operations. It ensures stability and consistency within the group's structure.

Here's what makes the identity element indispensable:
  • For an element \(e\) to be the identity in a group, it must satisfy: \(a * e = e * a = a\) for any element \(a\) in the group.
  • The identity element acts like a "do nothing" element, allowing other elements to retain their identity. For example, \(x * e = x\) and \(e * y = y\).
  • This property plays a crucial role in solving equations. When we simplify \(x * e = b * a^{-1}\), we conclude \(x = b * a^{-1}\) because the identity does not alter \(x\).
Understanding the identity element helps us see how elements interact in a group without changing their essence. It is a cornerstone for solving group-related problems.

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Most popular questions from this chapter

Untersuchen Sie die folgenden inneren Verkn眉pfungen \(\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}\) auf Assoziativit盲t, Kommutativit?t und Existenz von neutralen Elementen. (a) \((m, n) \mapsto m^{n}\). (b) \((m, n) \mapsto \operatorname{kgV}(m, n)\). (c) \((m, n) \mapsto \operatorname{gg} \mathrm{T}(m, n)\). (d) \((m, n) \mapsto m+n+m n\).

(a) \(2 X^{4}-3 X^{3}-4 X^{2}-5 X+6\) durch \(X^{2}-3 X+1\). (b) \(X^{4}-2 X^{3}+4 X^{2}-6 X+8\) durch \(X-1\).

Es seien die Abbildungen \(f_{1}, \ldots, f_{6}: \mathbb{R} \backslash\) \(\\{0,1\\} \rightarrow \mathbb{R} \backslash\\{0,1\\}\) definiert durch: $$ \begin{array}{ll} f_{1}(x)=x, & f_{2}(x)=\frac{1}{1-x}, \quad f_{3}(x)=\frac{x-1}{x} \\ f_{4}(x)=\frac{1}{x}, & f_{5}(x)=\frac{x}{x-1}, \quad f_{6}(x)=1-x \end{array} $$ Zeigen Sie, dass die Menge \(F=\left\\{f_{1}, f_{2}, f_{3}, f_{4}, f_{5}, f_{6}\right\\}\) mit der inneren Verkn眉pfung \(\circ:\left(f_{i}, f_{j}\right) \mapsto f_{i} \circ f_{j}\), wobei \(f_{i} \circ f_{j}(x)=f_{i}\left(f_{j}(x)\right)\), eine Gruppe ist. Stellen Sie eine Verkn眉pfungstafel f眉r \((F, \circ)\) auf.

ein kommutativer Erweiterungsring mit 1 von \(R[X]\) ist \(-\) der Ring der formalen Potenzreihen oder k眉rzer Potenzreihenring 眉ber \(R\). Wir schreiben \(P=\sum_{i \in \mathbb{N}_{0}} a_{i} X^{i}\) oder \(\sum_{i=0}^{\infty} a_{i} X^{i}\) (also \(P(i)=a_{i}\) ) f眉r \(P \in R[[X]]\) und nennen die Elemente aus \(R[[X]]\) Potenzreihen. Zeigen Sie au脽er- \(-\) dem: (a) \(R[[X]]\) ist genau dann ein Integrit盲tsbereich, wenn \(R\) ein Integrit盲tsbereich ist. (b) Eine Potenzreihe \(P=\sum_{i \in \mathbb{N}_{0}} a_{i} X^{i} \in R[[X]]\) ist genau dann invertierbar, wenn \(a_{0}\) in \(R\) invertierbar ist. (c) Bestimmen Sie in \(R[[X]]\) das Inverse von \(1-X\) und \(1-X^{2}\)

G. Zeigen Sie: (a) Es ist \(U_{1} \cup U_{2}\) genau dann eine Untergruppe von \(G\), wenn \(U_{1} \subseteq U_{2}\) oder \(U_{2} \subseteq U_{1}\) gilt. (b) Aus \(U_{1} \neq G\) und \(U_{2} \neq G\) folgt \(U_{1} \cup U_{2} \neq G\). (c) Geben Sie ein Beispiel f眉r eine Gruppe \(G\) und Untergruppen \(U_{1}, U_{2}\) an, sodass \(U_{1} \cup U_{2}\) keine Untergruppe von \(G\) ist.
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