# If H is the subgroup in Z 12, then H2 = (a) 5(b) 14 (c) {3, 6, 9, 0}  {2}(d) {5, 8, 11, 2} (e) {5, 6, 9, 0}(f) {3, 2, 5}

## Presentation on theme: "If H is the subgroup in Z 12, then H2 = (a) 5(b) 14 (c) {3, 6, 9, 0}  {2}(d) {5, 8, 11, 2} (e) {5, 6, 9, 0}(f) {3, 2, 5}"— Presentation transcript:

If H is the subgroup in Z 12, then H2 = (a) 5(b) 14 (c) {3, 6, 9, 0}  {2}(d) {5, 8, 11, 2} (e) {5, 6, 9, 0}(f) {3, 2, 5}

If H is the subgroup in Z 12, then H7 = (a) {3, 6, 9, 0}  {7}(b) {3, 7, 10} (c) {7, 10, 1, 4}(d) {7, 2, 9} (e) 10(f) 3  7

If H is the subgroup in S 3, then H(1,3) = (a) (1,2,1,3)(b) {(1,3,2), (1,3)} (c) {(1, 3, 2)}(d) {(1, 2, 3)} (e) {(1,2), (1,3)}(f) {(1,2,3), (1, 3)}

If H is the subgroup in S 3, then H(1,2) = (a) (1,2,1,2)(b) (2,4) (c) (1,2) 2 (d) {(1, 2)} (e) {(1,2), (2,1)}(f) H

If H is the subgroup in Z 4 x Z 6, then H(1,2) = (a) (2,1)(b) (3,4) (c) (2,2)  (1,2) (d) {(3,4), (1,0), (3,2), (1,4), (3,0), (1,2)} (e) {(1,2), (2,1), (3,4), (0,5), (1,1), (2,3)} (f) H

If H is the subgroup in Z 4 x Z 6, then H(2,0) = (a) (2,1)(b) (3,4) (c) (2,2)  (1,2) (d) {(3,4), (1,0), (3,2), (1,4), (3,0), (1,2)} (e) {(1,2), (2,1), (3,4), (0,5), (1,1), (2,0)} (f) H

How do we start this proof? (a) Assume A n is a subgroup of S n. (b)  (c) Assume a, b  H (d) Nonempty:

What is the next line of the proof? (a) Assume Ha = Hb (b) Assume ab -1  H (c) Assume a, b  H (d) Assume if ab -1  H then Ha = Hb.

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

What is the next line of the proof? (a) Assume Ha = Hb (b) Assume ab -1  H. (c) Assume a, b  H (d) Assume if ab -1  H then Ha = Hb.

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

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