Question

Let$\mathbf{N}$ be a subgroup of a group $\mathbf{G}$. Prove that$\mathbf{N}$ is normal if and only if$\mathbf{f}\left(N\right)\mathbf{=}\mathbf{N}$ for every inner automorphism$\mathbf{f}$ of $\mathbf{G}$.

Short answer

It is proved that$\mathrm{N}$ is the normal if and of only$\mathrm{f}\left(\mathrm{N}\right)=\mathrm{N}$ for every inner automorphism f of G.

Step-by-step solution

Prove that N is normal

Assume that$\mathrm{G}$be a group and$\mathrm{N}$be a subgroup of group, we need to show that$\mathrm{N}$is a normal subgroup if and only if$\mathrm{f}\left(\mathrm{N}\right)=\mathrm{N}$for every inner automorphism$\mathrm{f}$of $\mathrm{G}$.

Define the inner automorphism ${\mathrm{f}}_{\mathrm{c}}:\mathrm{G}\to \mathrm{G}$is an isomorphism of the form ${\mathrm{f}}_{\mathrm{c}}\left(\mathrm{g}\right)={\mathrm{cgc}}^{-1}$, for some $\mathrm{c}\in \mathrm{G}$and consider${\mathrm{f}}_{\mathrm{c}}\left(\mathrm{N}\right)=\mathrm{N}$, for all c belong to group G, then we have:

${\mathrm{cNc}}^{-1}=\mathrm{N}\forall \mathrm{c}\in \mathrm{G}$

This implies that$\mathrm{N}$ is the normal.

Prove that f(N)=N for every inner automorphism f of G

Consider that $\mathrm{N}$is normal subgroup of group $\mathrm{G}$, this implies that for $\mathrm{c}\in \mathrm{G}$, we have ${\mathrm{cNc}}^{-1}=\mathrm{N}$, which is equivalent to ${\mathrm{f}}_{\mathrm{c}}\left(\mathrm{N}\right)=\mathrm{N}$for every c belong to G.

This implies that $\mathrm{f}\left(\mathrm{N}\right)=\mathrm{N}$for every inner automorphism f of G.

Hence, it is proved that N is the normal if and of$\mathrm{f}\left(\mathrm{N}\right)=\mathrm{N}$ only for every inner automorphism f of G .