Homepage Solution manuals David S. Dummit Abstract Algebra Exercise 3.2.13 ($D_8$ as a subgroup of $S_4$)

Exercise 3.2.13 ($D_8$ as a subgroup of $S_4$)

Fix any labelling of the vertices of a square and use this to identify D 8 as a subgroup of S 4 . Prove that the elements of D 8 and ( 1 2 3 ) do not commute in S 4 .

Answers

Proof. We use the numbering of the vertices of a square given p.24, and We define D 8 = r , s , where r is the rotation of angle π 2 , and s the symmetry with respect to the x -axis. We use the numbering of the vertices of a square given p.24. Then D 8 acts on the set { 1 , 2 , 3 , 4 } by an action whose associate homomorphism φ : D 8 S 4 is defined by

φ ( r ) = ( 1 2 3 4 ) , φ ( s ) = ( 1 4 ) ( 2 3 ) .

Since an affine application of the plane is determined by three non-aligned points, this action is faithful, therefore

D 8 = r , s ρ , τ , where  ρ = ( 1 2 3 4 ) , τ = ( 1 4 ) ( 2 3 ) .

We identify D 8 with the subgroup H S 4 given by

( 1 2 3 4 ) , ( 1 2 ) ( 3 4 ) = { ( ) , ( 1 2 3 4 ) , ( 1 4 ) ( 2 3 ) , ( 1 3 ) ( 2 4 ) , ( 1 3 ) , ( 2 4 ) , ( 1 4 3 2 ) , ( 1 2 ) ( 3 4 ) } .

Let K = ( 1 2 3 ) S 4 . Since

( 1 2 3 4 ) ( 1 2 3 ) = ( 1 3 2 4 ) , ( 1 2 3 ) ( 1 2 3 4 ) = ( 1 3 4 2 ) ,

the permutations ( 1 2 3 4 ) H and ( 1 2 3 ) K do not commute in S 4 .

(But since H K = { 1 } , | 𝐻𝐾 | = | H | | K | = 8 3 = 24 = | S 4 | , so G = 𝐻𝐾 = 𝐾𝐻 .) □

User profile picture
2025-12-06 12:43
Comments