You want to verify that H = {even integers} is a subgroup of (ℤ, +). Which conditions do you actually need to check?
AAll four group axioms: closure, associativity, identity, and inverses
BNonemptiness, closure under addition, and closure under additive inverses
COnly closure under addition, since identity and inverses follow automatically from it
DOnly that H is a nonempty subset of G
Associativity is inherited for free because H ⊆ G — any triple a, b, c ∈ H satisfies (a+b)+c = a+(b+c) in ℤ, hence in H. Identity is automatic once you have closure under operation and inverses: if a ∈ H then a⁻¹ ∈ H, and a + a⁻¹ = 0 ∈ H by closure. So the subgroup test reduces to exactly three conditions: nonemptiness, closure under the operation, and closure under inverses. Option C is wrong because closure under addition alone does not guarantee inverses (ℤ⁺ is closed under addition but is not a subgroup).
Question 2 Multiple Choice
Is the set of positive integers ℤ⁺ a subgroup of (ℝ, +)?
AYes — it is nonempty and closed under addition
BNo — it is not closed under addition (e.g., 3 + 5 ∉ ℤ⁺)
CNo — it does not contain the identity element 0
DNo — it fails closure under inverses (the additive inverse of 3 is −3, which is not in ℤ⁺)
ℤ⁺ is closed under addition (option A's premise is correct) — that's the tempting mistake. But the subgroup test also requires closure under inverses. For any n ∈ ℤ⁺, its additive inverse −n is negative, so −n ∉ ℤ⁺. The test fails here. Note that option C (no identity) is a consequence of the same failure: once you know inverses aren't present, the identity can't be in H either, but the root failure by the subgroup test is the inverse condition.
Question 3 True / False
If H is a nonempty subset of a group G that is closed under the group operation and closed under taking inverses, then the identity element of G must also be in H.
TTrue
FFalse
Answer: True
Take any element a ∈ H (possible since H is nonempty). By closure under inverses, a⁻¹ ∈ H. By closure under the operation, a · a⁻¹ = e ∈ H. So the identity is automatically present — you never need to check it separately. This is the elegance of the subgroup test: two conditions imply the third.
Question 4 True / False
To confirm that a subset H of a group G is a subgroup, you should verify most four group axioms — closure, associativity, identity, and inverses — because H might not inherit properties from G.
TTrue
FFalse
Answer: False
Associativity is always inherited for free. Since H ⊆ G uses the same operation as G, and associativity holds for all elements of G, it holds in particular for all elements of H. There is no way for associativity to fail in a subset of an associative group. The subgroup test requires only nonemptiness, closure under the operation, and closure under inverses — three conditions, not four.
Question 5 Short Answer
Why does the subgroup test not require checking associativity, even though associativity is one of the four group axioms?
Think about your answer, then reveal below.
Model answer: Associativity is inherited from the ambient group G. Since H ⊆ G and H uses the same binary operation, for any a, b, c ∈ H we also have a, b, c ∈ G, where (ab)c = a(bc) already holds. The same equation therefore holds in H. Associativity cannot fail in a subset of a group because it depends on the operation, not on which elements are present.
This insight is what makes the subgroup test economical. Rather than re-verifying all axioms, you exploit the fact that H 'borrows' associativity from G. Only the conditions that depend on which elements are in H — closure under the operation and closure under inverses — need to be checked. Identity then follows as a consequence of those two.