Proving $(G,*)$ is a Group: Hints and Tips for a Simple Group Theory Problem

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SUMMARY

The discussion focuses on proving that the structure $(G, *)$ forms a group under the given properties: associativity, existence of an identity element, and existence of inverses. The key steps involve demonstrating that for any element $a \in G$, the identity element $e$ satisfies $a * e = a$ and that for each $a$, there exists an inverse $b$ such that $a * b = e$. The hints provided guide users to utilize the associative property and the definitions of identity and inverses effectively.

PREREQUISITES
  • Understanding of group theory concepts, specifically the definitions of a group.
  • Familiarity with binary operations and their properties.
  • Knowledge of the associative property in mathematics.
  • Ability to manipulate algebraic expressions involving sets and operations.
NEXT STEPS
  • Study the proof structure for group properties in abstract algebra.
  • Learn about the implications of the identity element in group theory.
  • Explore examples of groups, such as integer addition and matrix multiplication.
  • Investigate the role of inverses in group operations and their significance.
USEFUL FOR

Students of abstract algebra, mathematicians interested in group theory, and anyone seeking to understand the foundational properties of groups in mathematics.

alexmahone
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Let $G$ be a set and $*$ a binary operation on $G$ that satisfies the following properties:

(a) $*$ is associative,

(b) There is an element $e\in G$ such that $e*a=a$ for all $a\in G$,

(c) For every $a\in G$, there is some $b\in G$ such that $b*a=e$.

Prove that $(G, *)$ is a group.

My attempt: I know we're supposed to show that $a*e=a$ for (b) and that $a*b=e$ for (c). But I can't figure out how.

Any suggestions? Hints only as this is an assignment problem.
 
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Alexmahone said:
Let $G$ be a set and $*$ a binary operation on $G$ that satisfies the following properties:

(a) $*$ is associative,

(b) There is an element $e\in G$ such that $e*a=a$ for all $a\in G$,

(c) For every $a\in G$, there is some $b\in G$ such that $b*a=e$.

Prove that $(G, *)$ is a group.

My attempt: I know we're supposed to show that $a*e=a$ for (b) and that $a*b=e$ for (c). But I can't figure out how.

Any suggestions? Hints only as this is an assignment problem.
Hint for part (a). Pick $a\in G$ and let $a*e=b$. We want to show that $b=a$.

We know that there is a $c\in G$ such that $c*a=e$. Thus $c*a*e=c*b$, which gives $e=c*b$. We also have $e=c*a$. Now can you see what to do?

Try part (b) with this approach.
 
Got it, thanks!
 

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