Question

If$\mathit{G}$ is a group, show thatrole="math" localid="1657368651273" $\mathbf{⟨}\mathit{e}\mathbf{⟩}\mathbf{}$androle="math" localid="1657369423456" $\mathit{G}$ are normal subgroups.

Short answer

It is proved that$\mathbf{⟨}\mathit{e}\mathbf{⟩}$and $\mathit{G}$ are normal subgroups.

Step-by-step solution

Prove that ⟨e⟩ is normal subgroup

Since it is given that $\mathit{G}$be a group. Assume that$\mathit{g}\mathbf{\in }\mathit{G}$,this implies that:

$\mathit{g}\mathbf{\left\{}\mathit{e}\mathbf{\right\}}\mathbf{=}\mathbf{\left\{}\mathit{g}\mathbf{\right\}}\mathbf{=}\mathbf{\left\{}\mathit{e}\mathbf{\right\}}\mathit{g}$

Therefore, we can say that $\mathbf{⟨}\mathit{e}\mathbf{⟩}$ is normal group.

Prove that Gis normal subgroup

Since it is given that $\mathit{G}$be a group. Assume that $\mathit{g}\mathbf{\in }\mathit{G}$,this implies that:

$\mathit{g}\mathit{G}\mathbf{=}\mathbf{\{}\mathit{g}\mathit{h}\mathbf{:}\mathit{h}\mathbf{\in }\mathit{G}\mathbf{\}}$

$\mathbf{=}\mathit{G}$

$\mathbf{=}\mathbf{\{}\mathit{h}\mathit{g}\mathbf{:}\mathit{h}\mathbf{\in }\mathit{G}\mathbf{\}}$

$\mathbf{=}\mathit{G}\mathit{g}$

Therefore, we can say that $\mathit{G}$is normal group.

Hence, it is proved that localid="1657369562773" $\mathbf{⟨}\mathit{e}\mathbf{⟩}$and $\mathit{G}$is a normal subgroup of$\mathit{G}$.