Question 2. Let G be a finite group, H < G, N 4G, and gcd(|H|,|G/N|) = 1. Prove that H < N.
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[Groups and Symmetries] How do you solve question 2, thanks. Second picture is definitions
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- 5. For any subgroup of the group , let denote the product as defined in Definition 4.10. Prove that corollary 4.19:1.Prove part of Theorem . Theorem 3.4: Properties of Group Elements Let be a group with respect to a binary operation that is written as multiplication. The identity element in is unique. For each, the inverse in is unique. For each . Reverse order law: For any and in ,. Cancellation laws: If and are in , then either of the equations or implies that .Exercises 8. Find an isomorphism from the group in Example of this section to the multiplicative group . Sec. 16. Prove that each of the following sets is a subgroup of , the general linear group of order over .
- Let H and K be subgroups of a group G and K a subgroup of H. If the order of G is 24 and the order of K is 3, what are all the possible orders of H?34. Suppose that and are subgroups of the group . Prove that is a subgroup of .13. Assume that are subgroups of the abelian group . Prove that if and only if is generated by
- Exercises 31. Let be a group with its center: . Prove that if is the only element of order in , then .Let G be an abelian group of order 2n, where n is odd. Use Lagranges Theorem to prove that G contains exactly one element of order 2.For an integer n1, let G=Un, the group of units in n that is, the set of all [ a ] in n that have multiplicative inverses. Prove that Un is a group with respect to multiplication. (Sec. 3.5,3,6, Sec. 4.6,17). Find an isomorphism from the additive group 4={ [ 0 ]4,[ 1 ]4,[ 2 ]4,[ 3 ]4 } to the multiplicative group of units U5={ [ 1 ]5,[ 2 ]5,[ 3 ]5,[ 4 ]5 }5. Find an isomorphism from the additive group 6={ [ a ]6 } to the multiplicative group of units U7={ [ a ]77[ a ]7[ 0 ]7 }. Repeat Exercise 14 where G is the multiplicative group of units U20 and G is the cyclic group of order 4. That is, G={ [ 1 ],[ 3 ],[ 7 ],[ 9 ],[ 11 ],[ 13 ],[ 17 ],[ 19 ] }, G= a =e,a,a2,a3 Define :GG by ([ 1 ])=([ 11 ])=e ([ 3 ])=([ 13 ])=a ([ 9 ])=([ 19 ])=a2 ([ 7 ])=([ 17 ])=a3.