(3) Let a = (1 5 8 7)(1 3 4 6) e S9. (a) (b) Is a an even or odd permutation? Justify your answer. Find a-1. Find |a|. Find o51 (d)

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[Groups and Symmetries] How do you solve Q3, thanks

Suppose that ¢ : G → G' is a group homomorphism and there is a group homomorphism
(1) (a)
V : G' → G such that oo = idG', where idG' is the identity map on G'. Show that o is injective.
(b)
given by
Show that the converse of part (a) is not true as follows: Consider the map ø : Z3 → S3
Ф(0) %— с, Ф(1) 3 (132), Ф(2) — (123).
%3D
Show that o is injective but has no left-inverse, i.e there does not exist a homomorphism ý : S3 → Z3 such
that o o = idz3, where idz, is the identity map on S3.
(2)
of order 5 is in H.
Let G be a group of order 100 that has a subgroup H of order 25. Prove that every element of G
(3) Let α-
(a)
(b)
(c)
(d)
(1 5 8 7)(1 3 4 6) e S9.
Is a an even or odd permutation? Justify your answer.
Find a-1.
Find |a|.
Find a51.
(4)
subgroup of order p and only one subgroup of order q, prove that G is cyclic.
Suppose G is a group with order pq, where p and q are distinct prime numbers. If G has only one
(5)
Show that in a group G of odd order, the equation x2
= a has a unique solution for all a e G.
Transcribed Image Text:Suppose that ¢ : G → G' is a group homomorphism and there is a group homomorphism (1) (a) V : G' → G such that oo = idG', where idG' is the identity map on G'. Show that o is injective. (b) given by Show that the converse of part (a) is not true as follows: Consider the map ø : Z3 → S3 Ф(0) %— с, Ф(1) 3 (132), Ф(2) — (123). %3D Show that o is injective but has no left-inverse, i.e there does not exist a homomorphism ý : S3 → Z3 such that o o = idz3, where idz, is the identity map on S3. (2) of order 5 is in H. Let G be a group of order 100 that has a subgroup H of order 25. Prove that every element of G (3) Let α- (a) (b) (c) (d) (1 5 8 7)(1 3 4 6) e S9. Is a an even or odd permutation? Justify your answer. Find a-1. Find |a|. Find a51. (4) subgroup of order p and only one subgroup of order q, prove that G is cyclic. Suppose G is a group with order pq, where p and q are distinct prime numbers. If G has only one (5) Show that in a group G of odd order, the equation x2 = a has a unique solution for all a e G.
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