9 . 9 = e. = 92 = e. 47. Let G be an abelian group. Let H be the subset of G consisting of the identity e together with all elements of G of order 2. Show that H is a subgroup of G. pf) Let G be an abelian and H= 39EG: 191=2 or g=e? [ Closed ness] Show) high, E H for any highz EH [ Check ] | high2 1 = 2 7 ( high. ) = e [ Identity ] Show ) eEH [ Inverse] show ) haGH for any hell. [ check] In-1= 2 = (hi)2 =e. Hence H is a subgroup of G.49. Find a counterexample of Exercise 47 with the hypothesis that G is abelian omitted. Let G = S3 = ] v, ( 1, 2), (1, 3), ( 2, 37, ( 1, 2, 37, ( 1, 3, 27 6 Then H = P V , ( 1 , 2 ) , ( 1 1 3 ) , ( 2 , 3 ) 6 L H is not a subgroup! To Prove it ! !47. 48. 49. Let G be an abelian group. Let H be the subset of G consisting of the identity e together with all elements of G of order 2. Show that H is a subgroup of G. Following up the idea of Exercise 47 | determine whether H will always be a subgroup for every abelian group G if H consists of the identity e together with all elements of G of order 3; of order 4. For what positive integers n will H always be a subgroup for every abelian group G, if H consists of the identity e together with all elements of G of order n? Compare with Exercise 54 | of Section 5 |CJ. Find a counterexample of Exercise 47 |2 with the hypothesis that G is abelian omitted