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**Unformatted text preview: **Math 150A - Fall 09 - Homework 7&8 Selected Solutions (Based on solutions prepared by Jeff Ferreira ) 2.10.1 Let G be the group of invertible real upper triangular 2 2 matrices. Determine whether or not the following conditions describe normal subgroups H of G . If they do, identify the quotient G/H . (a) a 11 = 1 Proof. This is a normal subgroup of G , since it is the kernel of the homomorphism G R defined by a 11 a 12 a 22 7 a 11 . By the first isomorphism theorem, the quotient is isomorphic to the image, and the image is R . (b) a 12 = 0 Proof. This condition does not define a normal subgroup. (c) a 11 = a 22 Proof. This is the kernel of the homomorphism G R defined by a 11 a 12 a 22 7 a 11 a- 1 22 . Again the image is R . (d) a 11 = a 22 = 1 Proof. This is the kernel of the homomorphism G R R defined by a 11 a 12 a 22 7 ( a 11 a- 1 22 , a 11 ) . This time the image is R R 2.10.4a Consider the presentation (1 . 17) of the symmetric group S 3 . Let H be the subgroup { 1 , y } . Com- pute the product sets (1 H )( xH ) and (1 H )( x 2 H ), and verify they are not cosets. Proof. 1 H = { 1 , y } xH = { x, xy } x 2 H = x 2 , x 2 y (1 H )( xH ) = x, xy, x 2 y, x 2 (1 H )( x 2 H ) = x 2 , x 2 y, yx 2 , yx 2 y = x 2 , x 2 y, xy, x So the product sets have too many elements to be cosets, since any coset has the same number of elements as a given subgroup, and no subgroup of S 3 has order 4. 2.10.7 Find all normal subgroups N of the quaternion group H , and identify the quotients H/N . Proof. By looking at orders of elements, we see that the quaternion group has one subgroup of order 2: { 1 } . And three subgroups of order 4: { 1 , i } , { 1 , j } , { 1 , k } . There are also the trivial subgroups { 1 } and H . All subgroups of the quaternion group are normal. The quotient of H by itself is trivial, the quotient of H by any of the normal subgroups of order 4 gives a group of order 2, and so must be isomorphic to C 2 . The quotient of H by the normal subgroup of order 2 gives a group of order 4 which contains no element of order 4, and so is isomorphic to the Klein four group. The quotient of H by the trivial subgroup is isomorphic to H itself....

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