Question

Let $\mathit{N}$be a normal subgroup of a group $\mathit{G}$. Prove that $\mathit{G}\mathbf{/}\mathit{N}$ is simple if and only if there is no normal subgroup $\mathit{K}$such that $\mathit{N}\mathbf{⊈}\mathit{K}\mathbf{⊈}\mathit{G}$. (Hint: Corollary 8.23 and Theorem 8.24)

Short answer

Answer:

$G/N$ is simple if and only if there is no normal subgroup $k$such that $N⊈K⊈G$.

Step-by-step solution

Referring to the Corollary 8.23

Corollary 8.23

Let $\mathit{N}$ be a normal subgroup of a group $\mathit{G}$and let$\mathit{K}$be any subgroup of$\mathit{G}$that contains$\mathit{N}$. Then, $\mathit{K}$ is normal in $\mathit{G}$ if and only if$\mathit{K}\mathbf{/}\mathit{N}$is normal in $\mathit{G}\mathbf{/}\mathit{N}$.

Proving that G/N is simple if and only if there is no normal subgroup K such that N⊈K⊈G

Assume that $G/N$is a simple group.

Therefore, we know that $G/N$cannot have a normal subgroup.

Assume $K$ is a subgroup of $G$such that $N\subseteq K$.

Then, $K/N\subset G/N$because $K$is a subgroup of group $G$.

Now, $g\in G/N$and $k\in K/N$, which implies $gk{g}^{-1}\in K$.

This can only be true if $K$is a normal subgroup of $G$.

As corollary, $K/N$is also a normal subgroup of $G/N$.

Therefore, our assumption is wrong.

Hence, it is proved that $G/N$ is simple if and only if there is no normal subgroup $K$ such that $N⊈K⊈G$.