practice_2_soln - Practice Problem Solutions 1 Despite its...

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Practice Problem Solutions 1. Despite its appearance, this is a problem dealing exclusively with cosets and not using Lagrange’s Theorem. We start with the following observation. Let a, b K and suppose that a ( H K ) = b ( H K ). Then a - 1 b H K H and so aH = bH . It follows that the map { a ( H K ) | a K } → { aH | a H K } a ( H K ) 7→ aH is well-defined, i.e. does not depend on the choice of coset representative a . Moreover, this map is one-to-one. For if a, b K and aH = bH then a - 1 b H and since a - 1 b K we have a - 1 b H K so that a ( H K ) = b ( H K ). It follows that the number of cosets in the set on the left above (which is [ K : H K ]) is less than or equal to the number of cosets in the set on the right above (which is [ H K : H ]). That is [ K : H K ] [ H K : H ] . 2. In a recent homework assignment, we showed that the subgroup G = { σ S 4 | σ (4) = 4 } is isomorphic to S 3 . There’s nothing particularly special about the integer 4, and the same technique can be used to show that the three subgroups { σ S 4 | σ ( i ) = i } , i = 1 , 2 , 3 are all also isomorphic to S 3 . 3. Let ( ab ) be a transposition in S n . If a = 1 or b = 1 then ( ab ) is already among (12) , (13) , . . . , (1 n ). So suppose that a, b 6 = 1. Then, since a 6 = b , it is trivial to verify that ( ab ) = (1 a )(1 b )(1 a ) . It follows that any transposition can be written using only the transpositions (12) , (13) , . . . , (1 n ), and since any permutation can be written as a product of transpositions, that any element of S n can be written using only the transpositions (12) , (13) , . . . , (1 n ). 4. (a) Suppose that σ S n is a 3-cycle. Then σ has order 3. If σ 6∈ H then, since σ = ( σ 2 ) 2 , we must have σ 2 6∈ H as well. That is, σH 6 = H and σ 2 H 6 = H . But H has index 2 and so only has two cosets in
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practice_2_soln - Practice Problem Solutions 1 Despite its...

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