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?

We are given a set \( S \) with a binary operation \( * \) defined by \( ab = a \) for all \( a, b \in S \), 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 \( ab = a \) for all \( a, b \in S \). To prove that the operation is associative, we must show that for all \( a, b, c \in S \), we have:

\[
(a * b) * c = a * (b * c)
\]

#### Left-hand side:

\[
(a * b) * c = a * c
\]
(because \( a * b = a \) by the definition of the operation).

#### Right-hand side:

\[
a * (b * c) = a * b
\]
(because \( b * c = b \) by the definition of the operation).

But we know that \( a * b = a \), so:

\[
a * (b * c) = a
\]

Thus, we have:

\[
(a * b) * c = a * (b * c)
\]

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

### 2. Identity element of the operation

An identity element \( e \in S \) must satisfy:

\[
a * e = a \quad \text{and} \quad e * a = a \quad \text{for all } a \in S
\]

From the definition of the operation \( a * e = a \) is always true, regardless of \( e \), because \( a * b = a \) for any \( b \in S \).

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

### Conclusion:

- The operation defined by \( ab = a \) is associative for any set \( S \).
- The operation has an identity element only if \( S \) is a singleton set, i.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 \( ab = b \) instead of \( ab = a \)?
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.

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 have identity?

This problem explores the binary operation defined by ab = a for all a, b in a set S. Learn how to prove the operation is associative and determine the conditions under which the set has an identity element. Ideal for students studying abstract algebra at the college level.

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