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If H is the subgroup <3> in Z 12 , then H2 = 5 (b) 14

If H is the subgroup <3> in Z 12 , then H2 = 5 (b) 14 {3, 6, 9, 0}  {2} (d) {5, 8, 11, 2} (e) {5, 6, 9, 0} (f) {3, 2, 5}. If H is the subgroup <3> in Z 12 , then H7 = (a) {3, 6, 9, 0}  {7} (b) {3, 7, 10} {7, 10, 1, 4} (d) {7, 2, 9} (e) 10 (f) 3  7.

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If H is the subgroup <3> in Z 12 , then H2 = 5 (b) 14

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  1. If H is the subgroup <3> in Z12, then H2 = • 5 (b) 14 • {3, 6, 9, 0}  {2} (d) {5, 8, 11, 2} • (e) {5, 6, 9, 0} (f) {3, 2, 5}

  2. If H is the subgroup <3> in Z12, then H7 = • (a) {3, 6, 9, 0}  {7} (b) {3, 7, 10} • {7, 10, 1, 4} (d) {7, 2, 9} • (e) 10 (f) 3  7

  3. If H is the subgroup <(1,2)> in S3, then H(1,3) = • (a) (1,2,1,3) (b) {(1,3,2), (1,3)} • {(1, 3, 2)} (d) {(1, 2, 3)} • (e) {(1,2), (1,3)} (f) {(1,2,3), (1, 3)}

  4. If H is the subgroup <(1,2)> in S3, then H(1,2) = • (a) (1,2,1,2) (b) (2,4) • (1,2)2 (d) {(1, 2)} • (e) {(1,2), (2,1)} (f) H

  5. If H is the subgroup <(2,2)> in Z4 x Z6, then H(1,2) = • (2,1) (b) (3,4) (c) (2,2)  (1,2) • (d) {(3,4), (1,0), (3,2), (1,4), (3,0), (1,2)} • {(1,2), (2,1), (3,4), (0,5), (1,1), (2,3)} • H

  6. If H is the subgroup <(2,2)> in Z4 x Z6, then H(2,0) = • (2,1) (b) (3,4) (c) (2,2)  (1,2) • (d) {(3,4), (1,0), (3,2), (1,4), (3,0), (1,2)} • {(1,2), (2,1), (3,4), (0,5), (1,1), (2,0)} • H

  7. How do we start this proof? • Assume An is a subgroup of Sn. •  • Assume a, b  H • Nonempty:

  8. What is the next line of the proof? • Assume Ha = Hb • Assume ab-1 H • Assume a, b  H • Assume if ab-1  H then Ha = Hb.

  9. What is the next line of the proof? • Assume Ha = Hb •  • Assume a, b  H • Assume if ab-1  H then Ha = Hb.

  10. What is the next line of the proof? • Assume Ha = Hb • Assume ab-1 H. • Assume a, b  H • Assume if ab-1  H then Ha = Hb.

  11. What is the next line of the proof? • Assume Ha = Hb (b) Then a = Hb • (c) Then b-1a  H (d) Let y  H • (e) Let y  Ha (f) Let y = ab-1

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