(February 2013) Let \(G\) be a group, let \(H\subseteq G\) be a subgroup, and let \(N\trianglelefteq G\) be a normal subgroup. Define \[ NH=\{xh:x\in N\text{ and }h\in H\}. \] Prove that \(NH\) is a subgroup of \(G\).
Solution
Proof. (Nonempty): Since N and H are both nonempty, we may choose x∈ N and h∈ H. Then xh∈NH , so NH⁄= ∅. (Closed under ∗): Given x,y ∈ NH , write x = n1h1 and y = n2h2, for some n1,n 2 ∈ N and h1,h 2∈H. Since N is a normal subgroup of G, we have h1N =Nh 1. Thus, there is some n3∈N such that h1n2 =n3h1. So xy = (n1h1)(n2h2) =n1(h1n2)h2 =n1(n3h1)h2 = (n1n3)(h1h2)∈NH, as desired, since n1n3∈N and h1h2∈H. (Closed under inverses): Given x∈NH , write x =nh with n∈N and h∈H. Since N is a normal subgroup of G, we have hN =Nh. Thus, there is some n1∈N such that nh =hn1. So x−1 = (nh)−1 = (hn1)−1 =n−1 1 h−1∈NH, as desired, since n−1 1 ∈N and h−1∈H. QED Alternate Proof. (Nonempty): Since N andH are both nonempty, we may choose x∈N andh∈H. Then xh∈NH , so NH⁄= ∅. (Closed under ∗): Given x,y ∈ NH , write x = n1h1 and y = n2h2, for some n1,n 2 ∈ N and h1,h 2∈H. Then xy = (n1h1)(n2h2) =n1(h1n2)(h−1 1 h1)h2 = ( n1(h1n2h−1 1 ) ) (h1h2). Nown1,n 2∈N, and hence h1n2h−1 1 ∈N since N is normal, and therefore n1(h1n2h−1 1 )∈N. Since h1h2∈H, then the above equation shows xy∈NH , as desired. (Closed under inverses): Given x∈NH , write x =nh with n∈N and h∈H. Then x−1 = (nh)−1 =h−1n−1 = (h−1n−1h)h−1. Nowh−1n−1h∈N since N is normal in G, and h−1∈H, so the above equation shows x−1∈NH , as desired. QED