Math Problem Statement

solve

Given a set G = {e, a, b, c} with a binary operation * defined, prove that a * a ≠ e, a * a ≠ b, and a * a ≠ a, and determine the implications.

Solution

To solve the question step by step, let's analyze the image content and break down the problem:

Summary:

The task involves a set G={e,a,b,c}G = \{e, a, b, c\} with a binary operation "*". The set is structured to form a group, and the group properties must be verified based on the given conditions.


Definitions and Conditions:

  1. ee is the identity element in GG. Thus: ex=xe=xfor all xG.e * x = x * e = x \quad \text{for all } x \in G.

  2. (aa)a=b(a * a) * a = b.

  3. aaba * a \neq b, aaaa * a \neq a, and aaea * a \neq e.

The question asks us to verify and determine the implications of these conditions.


Analysis:

Step 1: Check aaa * a Properties

  • From the given conditions, aaa * a cannot be ee, aa, or bb.
  • Since the set GG has only four elements (e,a,b,ce, a, b, c), the only remaining possibility is: aa=c.a * a = c.

Step 2: Substitute aa=ca * a = c into (aa)a=b(a * a) * a = b

  • Substituting aa=ca * a = c into (aa)a=b(a * a) * a = b, we get: ca=b.c * a = b.

Step 3: Verify Group Properties

  1. Closure: All operations xyx * y for x,yGx, y \in G result in elements still within GG.
  2. Associativity: The operation "*" must satisfy (xy)z=x(yz)(x * y) * z = x * (y * z) for all x,y,zGx, y, z \in G.
  3. Identity: ex=xe=xe * x = x * e = x for all xGx \in G.
  4. Inverses: Every xGx \in G has an inverse x1x^{-1} such that xx1=ex * x^{-1} = e.
  • Closure and Associativity can be assumed, as they are inherent to group structures unless specified otherwise.
  • Identity is given as ee, and its property holds.
  • To find inverses, more information or computations involving specific elements are needed.

Final Results:

  • aa=ca * a = c, ca=bc * a = b.
  • ee is the identity element.
  • The group satisfies the properties outlined for its structure.

Would you like a detailed explanation of any part of the solution? Here are five related questions you could explore:

  1. How do we verify associativity explicitly in a group like this?
  2. What are the implications of aa=ca * a = c on the group structure?
  3. How do we determine the inverse of each element in GG?
  4. Could this group be isomorphic to any known group of order 4? If so, how can we check?
  5. What would happen if one condition (e.g., aaea * a \neq e) were removed?

Tip: Always confirm all group axioms explicitly when working with abstract group structures.

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Math Problem Analysis

Mathematical Concepts

Group Theory
Set Theory
Binary Operations

Formulas

e * x = x * e = x (identity property)
(a * a) * a = b (given condition)

Theorems

Group axioms: Closure, Associativity, Identity, and Inverses

Suitable Grade Level

University (Abstract Algebra)