Final answer:
If K is a normal subgroup of group G, and H is any subgroup of G, then the product set HK is itself a subgroup of G. This is proven by showing closure under group operation, existence of identity element, and inverses for all elements of HK within the set.
Step-by-step explanation:
To show that if K is a normal subgroup of G, and H is also a subgroup of G, then the set HK is a subgroup of G, we need to verify that HK satisfies the subgroup criteria. Specifically, we need to show that it is closed under the group operation, contains the identity element, and every element has an inverse in HK.
Firstly, because K is a normal subgroup of G, for every g in G and every k in K, we have that gkg-1 is also in K. Now, consider any elements h1k1 and h2k2 from HK. Their product is (h1k1)(h2k2). Because K is normal, we can write k1h2 as some element k3h2, where k3 is in K. Therefore, (h1k1)(h2k2) = h1(k3h2)k2 = h1h2(k3k2) since the group operation is associative.
h1h2 is in H since H is a subgroup and hence closed under the group operation, and k3k2 is in K because K is closed. Therefore, this product is an element of HK, proving closure.
Next, the identity element e of G is in K since K is a subgroup, and also in H, thus e is in HK.
For inverses, take any element hk in HK. The inverse (hk)-1 is k-1h-1. Since H and K are subgroups, h-1 is in H and k-1 is in K. Since K is normal, for h-1 in H we can find a k'-1 in K such that k-1h-1 equals h-1k'-1 which is in HK, showing that every element has an inverse in HK.
Therefore, HK is a subgroup of G when K is a normal subgroup of G and H is a subgroup of G.