If H is a subgroup of G and K is a normal subgroup of G,prove that HK = KH
By definition, HK contains all products of the form hk, where h∈H, and k∈K.
Similarly, by definition, KH contains all products of the form kh, where k∈K, and h∈H.
Next we need to prove that every product hk∈HK is also a member of KH.
By definition of normalcy of K in G, for k∈K,
xkx-1 = k ∀x∈G.
Let h∈H, and k∈K.
hk = (hkh-1)h
=kh ∀h∈H and k∈K
so HK ⊆KH.
Similarly, prove that KH⊆HK.
Hence conclude that HK=KH.
To prove that HK = KH, where H is a subgroup of G and K is a normal subgroup of G, we need to show that every element in HK is in KH, and vice versa.
First, let's show that every element in HK is in KH.
Let h ∈ H and k ∈ K. Since K is a normal subgroup of G, we know that khk^(-1) ∈ K for all h ∈ H and k ∈ K.
Now let's consider an element x ∈ HK. By definition, x = hk for some h ∈ H and k ∈ K.
We can rewrite x as x = (hke), where e is the identity element in G and is in both H and K.
Using the fact that khk^(-1) ∈ K, we can rewrite x as x = (khk^(-1))k.
Since khk^(-1) ∈ K and k ∈ K, we have x = (khk^(-1))k ∈ KH.
Therefore, every element in HK is in KH.
Next, let's show that every element in KH is in HK.
Let k ∈ K and h ∈ H. Since H is a subgroup of G, we know that hk ∈ H for all k ∈ K and h ∈ H.
Now let's consider an element y ∈ KH. By definition, y = kh for some k ∈ K and h ∈ H.
Using the fact that hk ∈ H, we can rewrite y as y = k(hk).
Since hk ∈ H and k ∈ K, we have y = k(hk) ∈ HK.
Therefore, every element in KH is in HK.
Since we have shown that every element in HK is in KH, and every element in KH is in HK, we can conclude that HK = KH.
Hence, we have proven that if H is a subgroup of G and K is a normal subgroup of G, then HK = KH.
To prove that HK = KH, we need to show that every element in HK is also in KH and vice versa.
Let's start by proving that HK is a subset of KH.
To do this, we need to show that for every element h in H and every element k in K, the element hk is in KH.
Since K is a normal subgroup of G, it means that for every element k' in K and every element g in G, the element gk'g^(-1) is also in K.
Now, let's take an arbitrary element hk from HK, where h is in H and k is in K.
Since H is a subgroup of G, it means that for every element h' in H and every element g in G, the element gh'g^(-1) is also in H.
Now, let's consider the element khk^(-1). Since k is in K, we can rewrite this as k(hk^(-1)).
To show that khk^(-1) is in KH, we need to find an element k' in K and an element h' in H such that khk^(-1) = k'h'.
Let's choose h' = k^(-1)hk. Since H is a subgroup of G, we know that h' is in H.
Now, let's consider k'. We'll choose k' = khk^(-1). Since K is a normal subgroup of G, it means that k' is in K.
With these choices of h' and k', we can see that khk^(-1) = k'h'.
Therefore, we have shown that for every element hk in HK, there exists an element k' in K and an element h' in H such that khk^(-1) = k'h'. This proves that HK is a subset of KH.
To prove that KH is also a subset of HK, we can follow a similar argument and choose k' = hkh^(-1) for every element kh in KH.
Therefore, we have shown that HK is a subset of KH and KH is a subset of HK, which implies that HK = KH.
In summary, we have proven that if H is a subgroup of G and K is a normal subgroup of G, then HK = KH.