If H is a subgroup of G, and K is a normal subgroup of G, prove that:
HK = KH
To prove that HK = KH, we need to show that every element in the set HK is also in the set KH, and vice versa.
Let's start by proving that HK is a subset of KH.
1. Take an arbitrary element h₁k₁ from HK, where h₁ ∈ H and k₁ ∈ K.
2. Since K is a normal subgroup of G, we know that for any k ∈ K and g ∈ G, gkg⁻¹ ∈ K.
3. Consider an arbitrary element g₁ ∈ G.
4. Then, (g₁h₁)k₁(g₁h₁)⁻¹ = g₁h₁k₁h₁⁻¹g₁⁻¹ = g₁h₂k₂, where h₂ = h₁k₁h₁⁻¹ ∈ H and k₂ = h₁k₁ ∈ K.
5. Therefore, we have shown that h₁k₁ ∈ KH, so HK ⊆ KH.
Now, let's prove that KH is a subset of HK.
1. Take an arbitrary element k₂h₂ from KH, where k₂ ∈ K and h₂ ∈ H.
2. Since H is a subgroup of G, we know that for any h ∈ H and g ∈ G, ghg⁻¹ ∈ H.
3. Consider an arbitrary element g₂ ∈ G.
4. Then, k₂(h₂g₂)k₂⁻¹ = k₂h₃k₃, where h₃ = g₂k₂g₂⁻¹ ∈ H and k₃ = k₂ ∈ K.
5. Therefore, we have shown that k₂h₂ ∈ HK, so KH ⊆ HK.
Since we have shown both HK ⊆ KH and KH ⊆ HK, we can conclude that HK = KH.