Let 7 and σ be elements of S5 such that: o(1) 2,0 (2) = 1, 0(3) = 4,0 (4) = 5,0 (5) = 3; o(1) = 5,0 (2) = 2, o(3) = 4, 0(4) = 3,0 (5) = 1. Calculate: (a) 100-100 (b) (0-172)33 (c) The order of the cyclic subgroup of S5 generated by 27³. (The order of a group is the number of elements in the group.)

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A: True or False   The subgroups of Z under addition are the groups nZ under addition for n.

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A:

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A: We use the Cayley table.

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A:

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A: Let’s assume that the H is a subgroup of S5. So,

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A: Prove that a group of class equation 24 = 1+1+4+4+4+4+6 does not have a normal subgroup of order 4.

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A: Let H be a subgroup of S5. we see that, HA5=HA5H∩A5 It follows that, A5≤HA5H∩A5≤S5 Since, S5=2A5, so…

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A: Subgroup

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A:

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A: Epimorphism: A homomorphism which is surjective is called Epimorphism.

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A: No of automorphism of Klein's 4-group:  K4={e,a,b,c} f1=eabceabc = I f2=eabceacb = bc f3=eabcecba =…

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A: z6={0,1,2,3,4,5}

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A: C* is group of non-zero comples numbers with multiplication

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A: Result: Let G, * be a group and H is a subset. The subset H is said to be subgroup, if for every a,…

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A: Consider the groups ℤ4  and  ℤ2⊕ℤ2. Clearly, both of the above groups are p-groups of order 4.

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