Math Problem Statement

Let S be a set. Prove that the law of composition defined by ab=a for all a and b in S is associative. For which sets does this law ha identity?

Solution

We are given a set SS with a binary operation * defined by ab=aab = a for all a,bSa, b \in S, and we are asked to:

  1. Prove that the operation is associative.
  2. Determine if this operation has an identity element and for which sets this law has one.

1. Prove that the operation is associative

The operation * is defined by ab=aab = a for all a,bSa, b \in S. To prove that the operation is associative, we must show that for all a,b,cSa, b, c \in S, we have:

(ab)c=a(bc)(a * b) * c = a * (b * c)

Left-hand side:

(ab)c=ac(a * b) * c = a * c (because ab=aa * b = a by the definition of the operation).

Right-hand side:

a(bc)=aba * (b * c) = a * b (because bc=bb * c = b by the definition of the operation).

But we know that ab=aa * b = a, so:

a(bc)=aa * (b * c) = a

Thus, we have:

(ab)c=a(bc)(a * b) * c = a * (b * c)

This shows that the operation is associative for all a,b,cSa, b, c \in S.

2. Identity element of the operation

An identity element eSe \in S must satisfy:

ae=aandea=afor all aSa * e = a \quad \text{and} \quad e * a = a \quad \text{for all } a \in S

From the definition of the operation ae=aa * e = a is always true, regardless of ee, because ab=aa * b = a for any bSb \in S.

However, ea=ee * a = e by the definition of the operation, but for ee to be an identity element, we require ea=ae * a = a. Thus, e=ae = a for all aSa \in S, which implies that SS must contain only one element. In other words, S={e}S = \{e\} for some element ee.

Conclusion:

  • The operation defined by ab=aab = a is associative for any set SS.
  • The operation has an identity element only if SS is a singleton set, i.e., S={e}S = \{e\}.

Would you like further details on any part of this explanation?

Here are some related questions:

  1. How do we determine if an operation is commutative?
  2. Can an operation have multiple identity elements?
  3. What are some examples of non-associative operations?
  4. How would the proof change if ab=bab = b instead of ab=aab = a?
  5. Can we extend this operation to infinite sets?

Tip: In associative operations, the grouping of elements doesn’t affect the result, but commutative operations require order to not matter either.

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Math Problem Analysis

Mathematical Concepts

Abstract Algebra
Binary Operations
Associativity
Identity Element

Formulas

(a * b) * c = a * (b * c)
a * e = a
e * a = a

Theorems

Associative Property
Identity Element Condition

Suitable Grade Level

College-Level Abstract Algebra