Question

Let H be a group and K a group with an action on H; i.e. with a specified homomorphism of K into the automorphism group of H. 24 Finite Groups Denote the image of an element h of H under the element k of K by h k

. Let G be the set of formal products kh with h∈H and k∈K, where 1 K
​
⋅h is taken to be h and k⋅1 H
​
is taken to be k. Define a product on G by setting kh⋅k ′
h ′
=(kk ′
)(h k ′
h ′
) for h and h ′
in H and k and k ′
in K. Prove that G becomes a group, H a normal subgroup of G and K a subgroup of G. Also G=KH=HK, H∩K=⟨1⟩ and h k
is just the conjugate k −1
hk of h by k in G. (This group G is called the split extension or semidirect product of H by K.)

Answer 1

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.