with details plz Let \( H \) be a subgroup of \( G \). Define the normalizer of \( H \) in \( G \) to be the subset\[ N_G(H) = \{ g \in G \mid gHg^{-1} = H \}. \](i) Prove that \( N_G(H) \) is a subgroup of \( G \) that contains \( H \).(ii) Prove that \( H \trianglelefteq N_G(H) \).(iii) Prove that if \( H \leq K \leq G \), and \( H \trianglelefteq K \), then \( K \subseteq N_G(H) \).

Let H be a subgroup of G. The normalizer of H in G is given by NGH= g∈G gHg-1=H A nonempty subset H of a group G is a subgroup of G if and only if (a) for all a,b∈H , a*b∈H (Closed under group operation) (b) for all a∈H, a-1∈H. (Inverse element exist) A subgroup H of G is said to be a normal subgroup if, for all g∈G, h∈H, ghg-1∈H (i) It is required to prove that NGH is a subgroup of G that contains H. Trivially e-1H e=H . so e∈NGH. Thus NGH is nonempty (a) Given f,g∈NGH, using associativity and the property of elements of NGH: fg-1Hfg=g-1f-1Hfg=g-1f-1Hfg=g-1Hg=H Therefore fg∈NGH Hence NGH is closed under group operation. (b) Given g∈NGH, g-1Hg=H so H=eHe=gg-1Hgg-1 =gg-1Hgg-1 =gHg-1 =g-1-1Hg-1Therefore g-1∈NGH From (a) and (b), NGH is a subgroup of G. Next to show that H⊂NGH Given a∈H it is required to show that a∈NGH (i.e) a-1Ha=H Given a-1ha∈a-1Ha, since H is closed under group operation, a-1ha∈H⇒a-1Ha⊂H Given h∈H, Since H is closed under group operation, aha-1∈H so h=a-1aha-1a∈ a-1Ha ⇒ H⊂ a-1Ha Hence a-1Ha=H (i.e) a∈NGH Thus a∈H⇒a∈NGH So H⊂NGH Thus it is proved that NGH is a subgroup of G that contains H. (ii) It is required to prove that H⊲ NGH, (i.e) H is a normal subgroup of NGH Given g∈NGH, h∈H : by the property of NGH, g-1Hg ⊂ H. Thus since g-1hg ∈ g-1Hg ⊂ H Hence g-1hg∈H Therefore H⊲ NGH (iii) Given that H is a subgroup of K and K is a subgroup of G and H⊲ k then for all k∈K, h∈H k h k-1∈H ..............(1) Then it is required to prove that K ⊂ NGH Given k,k-1∈K(since K is a subgroup of G) and h∈H, and from (1) k h k-1∈H ⇒kHk-1⊂ H ..............(2) And H=e H e =kk-1Hkk-1 =kk-1Hkk-1⊂ kHk-1Henc H⊂ kHk-1 .............(3) From (2) and (3), kHk-1=H Therefore, k∈NGH Thus k∈K⇒k∈NGH So K ⊂ NGH \