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ReviewExam2

  1. Definitions
  2. Two groups (G, *) and (G, *) are isomorphic if there exist a one-to-one correspondence i.e. , such that .
  3. Let H be subgroup of G. Then H is called normal subgroup of G if the left coset of H in G is a right coset of H in G OR
  4. The group of cosets of a normal subgroup H of the group G is called the quotient group or the factor group of G by H and it’s denoted by which G modulo H OR G mod H.

2. Solutions

  1. {(1), (123), (231), (12)} Not a subgroup because (123)(12) = (13) which does not belong to the set.
  2. Subgroup must contain the identity element
  3. {(1), (453), (12), (12)(453)}

(453)(12)= (12) (453)

(12)(453)(12) = (453)

This is a subgroup of S5

  1. {(1), (1234), (13)(24), (1432)} Yes, this is a subgroup of S5 since any multiplication of any element in the set belongs to the set

3. Solutions

  1. (1435)-1 = (1534)

(1534)(12) = (25413) = (23)(21)(24)(25)

  1. (153)(24) = (13)(15)(24)

4. Solutions

  1. (12345) = order 5
  2. (12)(234)(13) = (41)(32) = order lcm(2,2) = 2
  3. (1325)(46) = order lcm(4,2) = 4

5. Solutions

α = (1324) order of α = 4

α86 = α2

(1324)(1324) = (12) (34)

6. Solutions

Z4 under addition is isomorphic to {1, -1, i, -i} under multiplication

7. Solutions

U(14) = {1, 3, 5, 9, 11, 13} All elements relatively prime to 14

U(18) = {1, 5, 7, 11, 13, 17} All elements relatively prime to 18

Isomorphic

8. Solutions

Every finite group is isomorphic to a group of permutations.

9. Solutions

  1. S5 = 5! = 120
  2. A6 = ½ x6! = 360
  3. D5 = 10

10. Solutions

<4> = {0, 4, 8, 12}            Z16 = {0, 1, 2, 3, 4… 15}

0 = {0, 4, 8, 12}

1 = {1, 5, 9, 13}

2 = {2, 6, 10, 14}

3 = {3, 7, 11, 15}

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11. Solutions

Z16/<4> ≈ Z4 = {0, 1, 2, 3}

0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3 3 0 1 2

 

12. Solutions

Order (H) = 20 and Order (K) = 9. Then Order(HK) = LCM (20, 9) = 180.

 

13. Solutions

 

<(123)> = {(1), (123), (132)}

 

Left Cosets

(12)(123) = (23)

(13)(123) = (12)

(23)(123) = (13)

(123)(123) = (132)

(132)(123) = (1)

(12)(132) = (13)

(13)(132) = (23)

(23)(132) = (12)

(132)(132) = (123)

(123) (132) = (1)

 

 

Right Cosets


(123)(12) = (13)

(123)(13) = (23)

(123)(23) = (12)

(123)(123) = (132)

(123)(132) = (1)

(132)(12) = (23)

(132)(13) = (12)

(132)(23) = (13)

(132)(132) = (123)

(132)(123) = (1)


Yes it is a normal subgroup of S3 since distinct left cosets = distinct right cosets

14. Solutions

If H is a subgroup of a finite group G, then divides

15. Solutions

<a3> = {e, a3, a6, a9, a12}

<a> = {e, a, a2, a3, a4,…,a14}

Left Cosets

{e, a3, a6, a9, a12}

{a, a4, a7, a8, a14}

{a2, a5, a8, a9, a14}

16. Solutions

Order of U(40) = 16

Order of U(40)/K = 4

Since (4, 3) are relatively prime, the order of 3K = 4

17. Solutions

Order (S5) = 120

Order (H) = 3

Index = Order (S5)/Order (H) = 120/3= 40