Closure, Semigroup and Monoid: Worked Algebraic Structure Examples

Closure, associativity, semigroup and monoid are easy to memorise as separate definitions. The trouble starts when a question changes the set or operation, and the entire classification shifts. Use one order of checks and one operation table to test a candidate set-operation pair through the monoid conditions. A one-operation algebraic structure uses a single binary operation. Group Theory and Algebraic Structures: Groups, Rings and Fields Explained continues from monoids to groups, rings and fields; first, test closure, associativity and a two-sided identity in full.

1. Algebraic structures: move from closure to monoid in the right order

Start with a non-empty set S and a rule *. It is a binary operation on S only when every pair in S × S produces an element of S. This closed pair (S, *) is commonly called a magma in the one-operation hierarchy. A closed associative operation forms a semigroup. A semigroup with a two-sided identity forms a monoid.

The implications run one way. Every monoid is a semigroup, and every semigroup has a closed associative operation. Each converse needs extra evidence. An inverse for every element turns the monoid into a group.

There is one terminology detail. If a question declares *: S × S → S, closure is built into “binary operation”. If it merely gives a formula, you must still test closure.

2. Closure: test the result against the stated set

Closure requires a * b ∈ S for every a, b ∈ S. One successful pair proves nothing. One counterexample disproves closure.

Take S = {0, 1, 2, 3} under addition modulo 4. Here 3 ⊕ 2 = 1, which returns to S. A complete proof must cover arbitrary elements or every cell of the finite table.

Now take the odd integers under ordinary addition. They are not closed because 3 + 5 = 8, and 8 is not odd. All integers are closed under addition because the sum of any two integers is again an integer.

For a finite table, scan every body entry. Each must appear among the row and column labels. A table on {0, 1, 2, 3} containing 4 is not closed unless the stated rule reduces it, as modulo 4 does.

3. Associativity: brackets change, operand order does not

Associativity means (a * b) * c = a * (b * c) for every triple a, b, c ∈ S. Closure alone does not guarantee it. Ordinary subtraction is closed on the integers, yet

(5 - 3) - 1 = 2 - 1 = 1,

while

5 - (3 - 1) = 5 - 2 = 3.

Associativity changes the brackets but keeps a, b and c in the same order. Commutativity asks whether a * b = b * a, so a calculation that swaps operands does not prove associativity.

The proof burden also matters. One failed triple disproves associativity, while a few successful triples do not prove it. A four-element set has 4³ = 64 ordered triples. If the operation comes from a known associative law, prove it algebraically for arbitrary elements instead of sampling convenient cases.

4. Worked example: classify addition modulo 4 step by step

Let S = {0, 1, 2, 3} and define a ⊕ b = (a + b) mod 4. Its complete Cayley table is:

All 16 body entries are in S, so closure holds.

For arbitrary a, b, c ∈ S,

(a ⊕ b) ⊕ c = (((a + b) mod 4) + c) mod 4 = (a + b + c) mod 4,

and

a ⊕ (b ⊕ c) = (a + ((b + c) mod 4)) mod 4 = (a + b + c) mod 4.

Thus ⊕ is associative. For (3, 2, 1):

(3 ⊕ 2) ⊕ 1 = 1 ⊕ 1 = 2,

3 ⊕ (2 ⊕ 1) = 3 ⊕ 3 = 2.

Row 0 proves 0 ⊕ a = a, and column 0 proves a ⊕ 0 = a. Hence 0 is a two-sided identity and (S, ⊕) is a monoid. Its inverse pairs are (0, 0), (1, 3) and (2, 2), so it is also a group, an extra property not needed for the monoid test.

5. Semigroup without identity, and why the identity must be two-sided

Consider P = {1, 2, 3, ...} under ordinary addition. The sum of two positive integers is positive, so the operation is closed. Integer addition is associative. Therefore, (P, +) is a semigroup.

It is not a monoid. If e were an additive identity, a + e = a would force e = 0, but 0 ∉ P. Compare this with the non-negative integers N₀ = {0, 1, 2, ...}. Addition remains closed and associative, while 0 ∈ N₀ and 0 + a = a + 0 = a. Hence (N₀, +) is a monoid.

An identity must work on both sides, especially for a non-commutative operation. It is also unique. If e and f are both two-sided identities, then e * f = f because e is a left identity, and e * f = e because f is a right identity. Therefore, e = f.

One-sided identities are not enough. On S = {x, y}, define a * b = b. The operation is associative because both (a * b) * c and a * (b * c) equal c. Every element is a left identity because e * a = a, but no element is a right identity: x * y = y and y * x = x. This semigroup has no two-sided identity, so it is not a monoid.

6. How exams turn the definitions into short classification problems

Common tasks: classify a pair, find an identity, or disprove it at the first failure.

For integers, define a * b = a + b + 2. Closure holds. Also,

(a * b) * c = (a + b + 2) + c + 2 = a + b + c + 4,

a * (b * c) = a + (b + c + 2) + 2 = a + b + c + 4.

Thus * is associative. Also, a * e = a gives e = -2, and a * (-2) = (-2) * a = a. Therefore (ℤ, *) is a monoid.

Drill the table: check all 16 cells, explain why one triple is not a proof, find 0 from its row and column, then classify. Continue with Discrete Mathematics MCQs for mixed practice.

7. Closure, associativity and identity traps to avoid

The identity row proves e * a = a; its column proves a * e = a. A neat table does not prove associativity.

Check closure, associativity, then a two-sided identity. Stop at the first failure.

8. Closure, semigroup and monoid: the short version

Closure keeps every result inside the set. Associativity makes that closed structure a semigroup. A two-sided identity inside the set makes it a monoid. Inverses come later for a group. Remember S = {0, 1, 2, 3}, 3 ⊕ 2 = 1, and identity 0 as your memory check. For a structured route, use Engineering Mathematics for GATE Exam. Build the wider plan with GATE Guidance by Sanchit Sir and the GATE category.