Let G be a group let H be a subgroup of G and let N be a nor

Let G be a group, let H be a subgroup of G and let N be a normal subgroup of G. Prove that NH is a subgroup of G.

Solution

Clearly NH is nonempty, because e N H ==> e = e * e NH.

Let n, n\' N, and h, h\' H.
==> nh and n\'h\' NH.

(i) Closure under multiplication

(nh)(n\'h\') = n (hn\' h¹) hh\'.

Since N is normal in G and H is a subgroup of G, we have (hn\' h¹) N.

Hence, n (hn\' h¹) N, by closure in N.

Moreover, hh\' is in H, by closure of H.

Hence,

(nh)(n\'h\') = n (hn\' h¹) hh\' NH.

(ii) Inverse

Given that nh NH, we need to show that (nh)¹ = h¹ n¹ NH.

To show this, note that h¹ n¹ = (h¹ n¹ h) h¹

As before, (h¹ n¹ h) N, while h¹ H

Hence, h¹ n¹ = (h¹ n¹ h) h¹ NH.

Therefore, NH is a subgroup of G.