Question

Let N be a normal subgroup of G and let H be any subgroup of G. Let HN = hn. Show that HN is a subgroup of G, and is the smallest subgroup containing both N and H.

Answer 1

The question deals with proving that the set HN, formed by multiplying elements from a subgroup H and a normal subgroup N of a group G, is a subgroup of G and is the smallest containing both N and H. To prove this, one must verify that HN satisfies the criteria of containing the identity, being closed under multiplication, and containing inverses of its elements.

Step-by-step explanation:

The subject of the question involves group theory, which is a branch of abstract algebra in mathematics. Specifically, the question asks to prove that HN, the set formed by taking all possible products of elements from a subgroup H with elements from a normal subgroup N of a group G, is itself a subgroup of G.


Furthermore, it asks to show that HN is the smallest subgroup containing both N and H.

To prove that HN is a subgroup, we need to verify that it satisfies the subgroup criteria:

1. The identity of G is in HN.
   
   
2. If h1n1 and h2n2 are in HN, then their product h1n1h2n2 is also in HN.
   
   
3. If hn is in HN, then its inverse (hn)-1 is also in HN.

The criteria for the smallest subgroup are verified by the fact that any subgroup of G containing H and N must contain all the products h*n, since subgroups are closed under multiplication, and therefore must contain HN.