Consider H = {e, (12), (13), (23)} in S₃ — all transpositions plus the identity. Every element is its own inverse, so H is closed under inverses, and e ∈ H. Is H a subgroup of S₃?
AYes, because it contains the identity and all its own inverses
BNo, because it fails closure: (12)∘(13) = (132) ∉ H
CNo, because associativity doesn't hold inside H
DYes, because |H| = 4 divides |S₃| = 6
H fails the closure axiom. Even though every element is self-inverse and the identity is present, (12)∘(13) = (132) ∉ H, so H is not a subgroup. This is the classic trap: inverses + identity does not imply closure.
Question 2 Multiple Choice
In the one-step subgroup test, what does substituting a = b into the condition 'ab⁻¹ ∈ H for all a, b ∈ H' establish?
AThat the group operation is associative within H
BThat H is closed under the group operation
CThat the identity element e = aa⁻¹ belongs to H
DThat H is a normal subgroup of G
Setting a = b gives aa⁻¹ = e ∈ H, establishing that H contains the identity. From there, setting a = e (now known to be in H) shows b⁻¹ ∈ H (inverses), and replacing b with b⁻¹ recovers closure. One condition encodes all three axioms.
Question 3 True / False
If H is a nonempty subset of a group G satisfying ab⁻¹ ∈ H for all a, b ∈ H, then the identity element of G must belong to H.
TTrue
FFalse
Answer: True
Since H is nonempty, pick any a ∈ H. The condition with a = b gives aa⁻¹ = e ∈ H. The identity membership follows directly from the one-step test — you don't need to assume it separately.
Question 4 True / False
Any nonempty subset of a group that contains the identity element and is closed under taking inverses is a subgroup.
TTrue
FFalse
Answer: False
Closure under inverses and containing e are necessary but not sufficient. Closure under the group operation must also hold. The set {e, (12), (13), (23)} in S₃ has both properties yet is not a subgroup because it fails closure.
Question 5 Short Answer
Explain how the single condition 'ab⁻¹ ∈ H for all a, b ∈ H' in the one-step subgroup test encodes all three subgroup axioms: identity, inverses, and closure.
Think about your answer, then reveal below.
Model answer: Setting a = b gives e = aa⁻¹ ∈ H (identity). With e ∈ H, setting a = e gives b⁻¹ = eb⁻¹ ∈ H (inverses). Finally, replacing b with b⁻¹ in the original condition gives a(b⁻¹)⁻¹ = ab ∈ H (closure). Three axioms are recovered from one condition by strategic substitution.
The key is that the condition is assumed to hold for ALL pairs a, b ∈ H — so particular choices of a and b extract each axiom in turn. Understanding this shows why the test is logically equivalent to the full definition, not a shortcut that misses something.