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Abstract Algebra

Exercise 2.4.5 (The subgroup generated by two distinct elements of order $2$ in $S_3$ isf $S_3$)

Prove that the subgroup generated by two distinct elements of order 2 in S 3 is all of S 3 .

Answers

Proof. The elements of order 2 in S 3 are ( 1 2 ) , ( 2 3 ) , ( 1 3 ) .

Consider for instance H = ⟨ ( 1 2 ) , ( 2 3 ) ⟩ . Then the identity element ( ) = 1 S 3 ∈ H , and

( 1 2 3 ) = ( 1 2 ) ( 2 3 ) ∈ H , ( 1 2 3 ) 2 = ( 1 3 2 ) = ( 1 2 ) ( 2 3 ) ( 1 2 ) ( 2 3 ) ∈ H , ( 1 3 ) = ( 1 2 ) ( 2 3 ) ( 1 2 ) ∈ H .

Therefore G = H = ⟨ ( 1 2 ) , ( 2 3 ) ⟩ .

Any other pair of elements of order 2 is obtained by conjugation from this pair: for instance ( 1 3 ) = ( 3 1 ) = σ ( 1 2 ) σ − 1 , ( 1 2 ) = σ ( 2 3 ) σ − 1 , where σ = ( 1 3 2 ) . Then we obtain the same result for the pairs { ( 3 1 ) , ( 1 2 ) } and { ( 2 3 ) , ( 3 1 ) } .

S 3 = ⟨ ( 1 2 ) , ( 2 3 ) ⟩ , = ⟨ ( 3 1 ) , ( 1 2 ) ⟩ = ⟨ ( 2 3 ) , ( 3 1 ) ⟩ .

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2025-10-23 10:19
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