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Abstract Algebra

Exercise 2.4.8 ($S_4 = \langle (1\ 2\ 3\ 4), (1\ 2\ 4\ 3) \rangle$)

Prove that S 4 = ⟨ ( 1 2 3 4 ) , ( 1 2 4 3 ) ⟩ .

Answers

Proof. It is a well-known result that S n = ⟨ ( 1 2 ) , ( 1 2 3 … n ) (see Ex. 3.5.4 for a proof). So

S 4 = ⟨ ( 1 2 ) , ( 1 2 3 4 ) , ⟩ .

Put s = ( 1 2 3 4 ) and t = ( 1 2 4 3 ) , and H = ⟨ s , t ⟩ .

Then

s − 1 t s − 1 = ( 1 4 3 2 ) ( 1 2 4 3 ) ( 1 4 3 2 ) = ( 1 2 ) , (1)

so ( 1 2 ) ∈ H . Therefore

S 4 = ⟨ ( 1 2 3 4 ) , ( 1 2 ) ⟩ ⊆ H ,

where H is a subgroup of S 4 , hence H = S 4 .

S 4 = ⟨ ( 1 2 3 4 ) , ( 1 2 4 3 ) ⟩ .

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Note: I found (1) with Sagemath:

sage: G = PermutationGroup([[(1,2,3,4)], [(1,2,4,3) ] ])
sage: u = G(’(1,2)’);u
(1,2)
sage: u.word_problem(G.gens(), False)
(’x1^-1*x2^-3*x1^-1’, ’(1,2,3,4)^-1*(1,2,4,3)^-3*(1,2,3,4)^-1’)
sage: s , t = G.gens()[0], G.gens()[1]; s,t
((1,2,3,4), (1,2,4,3))
sage: s^-1 * t * s^-1
(1,2)

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2025-10-24 17:03
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