ReviewExam2
- Definitions
- Two groups (G, *) and (G^{’}, *^{’}) are isomorphic if there exist a one-to-one correspondence i.e. , such that .
- 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
- 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), (123), (231), (12)} Not a subgroup because (123)(12) = (13) which does not belong to the set.
- Subgroup must contain the identity element
- {(1), (453), (12), (12)(453)}
(453)(12)= (12) (453)
(12)(453)(12) = (453)
This is a subgroup of S_{5}
- {(1), (1234), (13)(24), (1432)} Yes, this is a subgroup of S_{5} since any multiplication of any element in the set belongs to the set
3. Solutions
- (1435)^{-1} = (1534)
(1534)(12) = (25413) = (23)(21)(24)(25)
- (153)(24) = (13)(15)(24)
4. Solutions
- (12345) = order 5
- (12)(234)(13) = (41)(32) = order lcm(2,2) = 2
- (1325)(46) = order lcm(4,2) = 4
5. Solutions
α = (1324) order of α = 4
α^{86} = α^{2}
(1324)(1324) = (12) (34)
6. Solutions
Z_{4} 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
- S_{5} = 5! = 120
- A_{6} = ½ x6! = 360
- D_{5} = 10
10. Solutions
<4> = {0, 4, 8, 12} Z_{16} = {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}
11. Solutions
Z_{16}/<4> ≈ Z_{4} = {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 S_{3} since distinct left cosets = distinct right cosets
14. Solutions
If H is a subgroup of a finite group G, then divides
15. Solutions
<a^{3}> = {e, a^{3}, a^{6}, a^{9}, a^{12}}
<a> = {e, a, a^{2}, a^{3}, a^{4},…,a^{14}}
Left Cosets
{e, a^{3}, a^{6}, a^{9}, a^{12}}
{a, a^{4}, a^{7}, a^{8}, a^{14}}
{a^{2}, a^{5}, a^{8}, a^{9}, a^{14}}
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 (S_{5}) = 120
Order (H) = 3
Index = Order (S_{5})/Order (H) = 120/3= 40