To prove that G is a group, we need to show that it satisfies the group axioms of closure, associativity, identity, and inverse.
Closure: For any kh and k′h′ in G, their product kh⋅k′h′ is also in G.
Associativity: The product operation in G is associative: (kh⋅k′h′)⋅k″h″ = kh⋅(k′h′⋅k″h″) for any kh, k′h′, and k″h″ in G.
Identity: The element 1K⋅1H serves as the identity element in G.
Inverse: For any kh in G, its inverse is given by k−1⋅h−1.
H is a normal subgroup of G because for any kh in G and any h′ in H, the conjugate k−1⋅hk is also in G.
K is a subgroup of G since it is closed under the product operation and has the identity element.
G = KH = HK, which means that every element in G can be expressed as a product of an element in K and an element in H.
H∩K = ⟨1⟩, meaning the intersection of H and K is the trivial subgroup consisting only of the identity element.
The expression hk represents the conjugate of h by k in G, denoted as k−1⋅hk.