Verifying ⟨(123)⟩ is a Subgroup of S₄

Abstract Algebra · Axiom Academy

EXAMPLE Verifying ⟨(123)⟩ is a Subgroup of S₄ Step-by-step verification using the subgroup test: compute powers, verify closure, and check inverses Excellent work! You've verified that ⟨(123)⟩ is a subgroup of S₄. Here's what we learned: Computing Powers: (123) has order 3, so (123)³ = e. The cyclic subgroup contains e, (123), (132) . Subgroup Test: To verify a subset is a subgroup, check: (1) contains identity, (2) closed under operation, (3) contains inverses. Closure Verification: We can verify closure by checking all pairwise products or by recognizing it's generated by a single element. Order Connection: The element (123) has order 3, and ⟨(123)⟩ has 3 elements. In cyclic groups, |⟨g⟩| = order(g). Inverse Relationships: Notice that (123) and (132) are inverses: (123) ∘ (132) = e. In cyclic groups, g⁻¹ = gⁿ⁻¹. This example demonstrates the finite subgroup test in action. Any cyclic subgroup ⟨g⟩ generated by an element of finite order is automatically a subgroup, which is why computing the element's order is often the quickest verification method!

This is the written version of the interactive lesson above. See the full Abstract Algebra course.