Question

Let N and K be subgroups of a group G. If N is normal in G, prove that $NK=\left\{nk|n\in N,k\in K\right\}$is a subgroup of G. [Compare Exercise 26 (b) of section 7.3]

Short answer

It is proved that $NK=\left\{nk|n\in N,k\in K\right\}$ is a subgroup of G.

Step-by-step solution

Referring to the Theorem 7.11:

Theorem 7.11:

A nonempty subset H of a group G is a subgroup of G provided that

(i) if $a,b\in H$, then $ab\in H$; and

(ii) if $a\in H$then $a^{-1}\in H.$

Proving that NK=nk|n∈N,k∈K  is a subgroup of G

To prove that $NK=\left\{nk|n\in N,k\in K\right\}$ is a subgroup of G, we must prove that it is closed under the inverse composition.

Suppose $n_1{k}_1,n_2{k}_2\in nk$

Then,$\left(n_1{k}_1\right),{\left(n_2{k}_2\right)}^{-1}=n_1{k}_1{k_2}^{-1}{n}_2$

Since $k_2{k_1}^{-1}{k}_1{k_2}^{-1}=1$, we can multiply it in the above equation

$\begin{array}{c}\left(\left(n_1{k}_1\right),{\left(n_2{k}_2\right)}^{-1}\right)=n_1{k}_1{k_2}^{-1}{n}_2\\ =n_1{k}_1{k_2}^{-1}{n}_2{k}_2{k_1}^{-1}{k}_1{k_2}^{-1}\end{array}$

In the above equation $k_1{k_2}^{-1}\in K$ (because K is a subgroup of G)

And $k_1{k_2}^{-1}{n}_2{k}_2{k_1}^{-1}\in N$, because it is a normal subgroup

So, from all the above, it can be concluded that

$n_1{k}_1{k_2}^{-1}{n}_2{k}_2{k_1}^{-1}{k}_1{k_2}^{-1}\in NK$

$\left(n_1{k}_1\right),{\left(n_2{k}_2\right)}^{-1}\in NK$

Since NK it is closed under the inverse, it is proved that $NK=\left\{nk|n\in N,k\in K\right\}$ is a subgroup of G.