Commutative Property and Abelian Groups: Tests, Worked Examples and Exam Traps

Commutativity looks like a simple swap rule, but that simplicity causes a common mistake: a commutative operation need not form a group, and a group need not be Abelian. Swapping operands, changing brackets, finding an identity and finding inverses are separate checks. A custom operation needs five property checks; a finite Cayley table can reveal commutativity, and one counterexample is enough to disprove it in S_3.

Commutative property versus Abelian group: the extra condition

A binary operation * on a set G is commutative when a*b = b*a for every a,b in G. A group already has closure, associativity, a two-sided identity and an inverse for every element. It is Abelian only when commutativity also holds. Every Abelian group is a group, but not every group is Abelian.

Swapping is different from regrouping. On the integers, define x ⊕ y = x^2 + y^2. The operation is commutative because x^2 + y^2 = y^2 + x^2. It is not associative. With x=1, y=2 and z=3:

• (1 ⊕ 2) ⊕ 3 = 5 ⊕ 3 = 25 + 9 = 34

• 1 ⊕ (2 ⊕ 3) = 1 ⊕ 13 = 1 + 169 = 170

One unequal triple disproves associativity. Several successful trials would not prove it. The GATE CS Exam Preparation page gives the broader catalog route.

Fully worked example: prove that shifted addition is an Abelian group

Let G=Z and define a*b = a+b+3. Symmetry is not enough, so check every property in order.

For inverse uniqueness, left- and right-sided group equations, and reversed product laws, use Inverse Property in Group Theory: Concepts and a Complete Worked Example. The calculation below treats inverses as one of five properties required for an Abelian-group classification.

1. Closure: If a,b are integers, then a+b+3 is also an integer.

2. Associativity:

• (a*b)*c = (a+b+3)+c+3 = a+b+c+6

• a*(b*c) = a+(b+c+3)+3 = a+b+c+6

Both parenthesisations give the same result.

1. Identity: Solve instead of guessing. From a*e=a, we get a+e+3=a, so e=-3. The other side also works: e*a = -3+a+3 = a.

2. Inverse: The inverse of a must combine with it to give e=-3. Solving a*b=-3 gives a+b+3=-3, hence b=-a-6, an integer for every integer a. It also works on the left. For a=4, its inverse is -10, and 4*(-10)=4-10+3=-3=(-10)*4.

3. Commutativity: For arbitrary integers, a*b = a+b+3 = b+a+3 = b*a.

All five checks pass, so (Z,*) is an Abelian group. Commutativity did not supply associativity, an identity or inverses. Every line of the audit was needed.

Finite Abelian example: read symmetry in the Cayley table of Z_4

Take Z_4={0,1,2,3} with addition modulo 4, written +_4. Its Cayley table is:

All entries lie in Z_4, so closure holds. Row 0 and column 0 reproduce the headers, making 0 the two-sided identity. The zero cells give the inverse pairs 0 <-> 0, 1 <-> 3 and 2 <-> 2.

Reflection across the main diagonal leaves the table unchanged. For example, row 1, column 3 and row 3, column 1 both contain 0. Row 2, column 3 and row 3, column 2 both contain 1. This symmetry shows a+_4 b = b+_4 a. A symmetric table proves commutativity for the displayed operation, but not associativity by itself. Here associativity follows from integer addition modulo 4. Thus Z_4 under addition modulo 4 is a finite Abelian group.

A valid group need not be Abelian: the S_3 counterexample

For finite-versus-infinite groups, cyclicity, and element-order classification, use Classification of Finite and Infinite Groups: Orders, Cyclicity and Worked Examples. The S_3 calculation here answers a narrower question: reversing two products changes the result, so commutativity fails.

S_3 is the group of six permutations of {1,2,3} under composition:

e, (12), (13), (23), (123), (132).

Use the convention that the right-hand permutation acts first. Let a=(12) and b=(23). For a o b=(12)o(23), track each input through (23) and then (12):

• 1 -> 2

• 2 -> 3

• 3 -> 1

Therefore a o b=(123). Reverse the order. The composition b o a=(23)o(12) maps:

• 1 -> 3

• 3 -> 2

• 2 -> 1

Therefore b o a=(132). Since (123) != (132), composition is not commutative in S_3. One pair disproves the universal law. S_3 is a non-Abelian group, showing that associativity, identity and inverses do not force commutativity. This does not make every permutation group non-Abelian; small or special ones can commute.

Classify familiar sets and operations without pattern matching

For binary strings over Sigma={0,1}, concatenation has the empty string epsilon as a two-sided identity. However, the string 0 has no inverse that can concatenate with it to produce epsilon. Concatenation is also noncommutative because 0 followed by 1 gives 01, while the reverse gives 10. Thus (Sigma*, concatenation) is a noncommutative monoid, not a group.

How exam-style questions test commutativity and Abelian groups

Most questions use one of four tasks: identify the extra Abelian condition, classify a familiar set-operation pair, expand a custom operation through every property, or read a Cayley or truth table for symmetry. Mirrored table entries test commutativity. Identity, inverses and associativity remain separate checks.

A useful proof starts with a group satisfying (ab)^2=a^2b^2 for every a,b. Expand it:

abab = aabb.

Left-multiply by a^-1 to obtain bab=abb. Then right-multiply by b^-1 to obtain ba=ab. Hence the group is Abelian. These cancellations are valid because inverses and associativity are already available in a group.

Practise all four task types, not only the one you find easiest. For timed mixed practice, use the GATE Test Series: Mocks and Topic-wise Tests.

Common traps and reliable proof moves

1. Treating examples as proof: To prove commutativity, work with arbitrary a,b. To disprove it, one counterexample such as (12)o(23) != (23)o(12) is enough.

2. Swapping instead of regrouping: Associativity compares (a*b)*c with a*(b*c). Write both before calculating.

3. Forgetting the set: Multiplication is commutative on Q, but (Q, multiplication) is not a group because 0 has no inverse. Removing 0 changes the answer.

4. Checking only one side: A convenient right identity or one inverse is insufficient. Verify both sides and cover every element.

5. Trusting table symmetry too soon: A symmetric Cayley table establishes commutativity, not all group axioms.

Use one fixed audit: set and operation, closure, associativity, two-sided identity, inverse for every element, then commutativity.

Short version and the next step

Commutativity means a*b=b*a. An Abelian group is a group that also satisfies this law. Shifted addition a*b=a+b+3 passed all five checks with identity -3, while S_3 failed commutativity for the two chosen transpositions.

Now rebuild the Z_4 table and redo the shifted-addition proof from a blank page. Then use GATE Guidance by Sanchit Sir as a sequenced route through Discrete Mathematics. Repeat the audit in order until no property is skipped.