Consider the set of odd integers under ordinary addition. Which group axiom fails?
AAssociativity — regrouping odd integers changes the sum
BClosure — the sum of two odd integers is even, which is not in the set
CIdentity — there is no odd integer that acts as the additive identity
DInverses — the additive inverse of an odd integer is not odd
The sum of any two odd integers is always even (e.g., 3 + 5 = 8), so the set of odd integers is not closed under addition. The result leaves the set, violating closure immediately. Note that options C and D are also technically true (0 is not odd; −3 is odd so inverses exist... wait, −3 is odd), but closure fails first and most directly. Closure is the axiom students most often overlook because it seems obvious — but it must be verified for the specific set, not just the operation.
Question 2 Multiple Choice
Which of the following is a group under its given operation?
C(ℚ \ {0}, ×) — nonzero rational numbers under multiplication
D(2ℤ, ×) — even integers under multiplication
The nonzero rationals under multiplication satisfy all four axioms: closure (product of two nonzero rationals is a nonzero rational), associativity (inherited from multiplication), identity (1 is a nonzero rational), and inverses (every p/q ≠ 0 has inverse q/p). The others fail: (ℤ, ×) has no inverse for 2 since 1/2 ∉ ℤ; (ℕ, +) has no inverse for 1 since −1 ∉ ℕ; (2ℤ, ×) has no identity since 1 is not even. Whether a structure is a group depends on the combination of set AND operation.
Question 3 True / False
The identity element in a group is unique — no group can have two different elements that both satisfy the identity axiom.
TTrue
FFalse
Answer: True
This is a provable theorem about groups: suppose e and e′ are both identity elements. Then e = e ∗ e′ (since e′ is an identity) = e′ (since e is an identity). So e = e′. The proof works for any group, no matter what the set or operation is — this is the abstraction payoff. You prove uniqueness once for abstract groups and it immediately applies to all groups: integer addition, rational multiplication, rotations, permutations, and every other example simultaneously.
Question 4 True / False
Whether a set forms a group depends mainly on the properties of the set, not on which binary operation is used.
TTrue
FFalse
Answer: False
Both the set and the operation together determine whether a group exists. The integers ℤ form a group under addition (every integer has an additive inverse −n) but not under multiplication (2 has no multiplicative inverse in ℤ). The set is the same; the operation changes everything. This is why the proper notation specifies both: (ℤ, +) is a group, (ℤ, ×) is not. Students who think of 'the integers' as simply being or not being a group are missing half the definition.
Question 5 Short Answer
Explain why (ℤ, ×) — the integers under multiplication — fails to be a group. Which axiom is violated, and why does it fail for this particular set-operation pair?
Think about your answer, then reveal below.
Model answer: The integers under multiplication fail the inverse axiom. The group axiom requires that for every element a in the set, there exists an element a⁻¹ in the same set such that a × a⁻¹ = 1. For a = 2, the multiplicative inverse would need to be 1/2 — but 1/2 is not an integer. The same problem occurs for every integer except 1 and −1 (which are their own inverses). Since most elements lack inverses within ℤ, (ℤ, ×) fails to be a group.
The other axioms do hold: closure (product of two integers is an integer), associativity (always true for multiplication), and identity (1 is the multiplicative identity in ℤ). The failure is specifically and only the inverse axiom. This is why the nonzero rationals (ℚ \ {0}, ×) form a group while integers do not: the rationals contain all the fractions needed to invert every nonzero element.