Answer:
Let (M,*) be a monoid, which means that M is a set equipped with an associative binary operation * and an identity element e such that for all a, b, and c in M:
a*(bc) = (ab)*c (associativity)
ae = ea = a (identity)
We want to show that the semigroup of subsets of M, denoted by (P(M),•), where • is the set intersection operation, is also a monoid.
First, let's show that (P(M),•) is a semigroup. To prove that (P(M),•) is closed under the operation •, we need to show that for any A, B in P(M), A • B is also in P(M). This is true since A • B is a subset of M. Moreover, the operation • is associative, which means that for any A, B, and C in P(M), we have:
(A • B) • C = A • (B • C)
Therefore, (P(M),•) is a semigroup.
Next, we need to show that (P(M),•) has an identity element. The identity element for the semigroup (P(M),•) is the set M itself, since for any A in P(M), we have:
M • A = A • M = A
Therefore, (P(M),•) has an identity element.
Finally, we need to show that (P(M),•) satisfies the associative and identity properties of a monoid. Since we have already shown that (P(M),•) is a semigroup with an identity element, we only need to show that • is associative and that M is the identity element.
Associativity: For any A, B, and C in P(M), we have:
(A • B) • C = A ∩ B ∩ C
A • (B • C) = A ∩ (B ∩ C)
Since the intersection operation is associative, we have:
A ∩ (B ∩ C) = (A ∩ B) ∩ C
Therefore, (A • B) • C = A • (B • C), and • is associative.
Identity: For any A in P(M), we have:
M • A = M ∩ A = A
A • M = A ∩ M = A
Therefore, M is the identity element for (P(M),•), and (P(M),•) is a monoid.
Hence, we have shown that the semigroup of subsets of a monoid is also a monoid.