Question

(Second Isomorphism Theorem) Let K and N be subgroups of a group G, with N normal in G. Then$\mathit{N}\mathit{K}\mathbf{=}\mathbf{\left\{}\mathbf{n}\mathbf{k}\mathbf{|}\mathbf{n}\mathbf{\in }\mathbf{k}\mathbf{,}\mathbf{k}\mathbf{\in }\mathbf{K}\mathbf{\right\}}$ is a subgroup of G that contains both K and N by Exercise 20 of Section 8.2.

Prove that N is a normal subgroup of NK.

Short answer

It is proved that, N is a normal subgroup of NK.

Step-by-step solution

Step by Step Solution Step 1: Definition of NK

$\mathit{N}\mathit{K}\mathbf{=}\mathbf{\left\{}\mathbf{n}\mathbf{k}\mathbf{|}\mathbf{n}\mathbf{\in }\mathbf{k}\mathbf{,}\mathbf{k}\mathbf{\in }\mathbf{K}\mathbf{\right\}}$is a subgroup of G that contains both N and K.

Proving that N is a normal subgroup in NK

As we know NK is a normal subgroup of G. Therefore, from the inverse property of the subgroup, we know that for any arbitrary $nk\in NK$, $\left(nk\right),{\left(nk\right)}^{-1}\in NK$.

To prove N is a normal subgroup of NK.

It is enough to show that for any arbitrary nk, $\left(nk\right)N{\left(nk\right)}^{-1}\subseteq N$.

Let’s consider $\left(nk\right)N{\left(nk\right)}^{-1}$

Since NK is a subgroup of G and $\left(nk\right),{\left(nk\right)}^{-1}\in NK$, $\left(nk\right)N{\left(nk\right)}^{-1}\in N$.

Therefore,$\left(nk\right)N{\left(nk\right)}^{-1}\in N$ implies $\left(nk\right)N{\left(nk\right)}^{-1}\subseteq N$.

Hence, it is proved that N is a normal subgroup of NK.