Question

Let$\mathbf{N}$be subgroup of a group$\mathbf{G}$ . Suppose that, for each $\mathbf{a}\mathbf{\in }\mathbf{G}$, there exists$\mathbf{b}\mathbf{\in }\mathbf{G}$ such that $\mathbf{Na}\mathbf{=}\mathbf{bN}$. Prove that $\mathbf{N}$ is a normal subgroup.

Short answer

It is proved that $\mathrm{N}$is a normal subgroup.

Step-by-step solution

Theorem

Use the statement, if$\mathrm{a}$ is an element of the group $\mathrm{G}$, then $\mathrm{aG}=\mathrm{G}$.

Prove

Assume that $\mathrm{a}$ be any arbitrary element of$\mathrm{G}$ and $\mathrm{b}\in \mathrm{G}$,$\mathrm{N}$ is a subgroup of group $\mathrm{G}$, this implies that $\mathrm{Na}=\mathrm{bN}$.

This implies that for $\mathrm{n}\in \mathrm{N}$, there exits${\mathrm{n}}_1\in \mathrm{N}$ such that $\mathrm{na}={\mathrm{n}}_1\mathrm{b}$.

Assume $\mathrm{n}=\mathrm{e}$, there exits${\mathrm{n}}_1\in \mathrm{N}$ such that $\mathrm{a}=\mathrm{ea}={\mathrm{bn}}_1$.

Therefore, we have:

$\begin{array}{rcl}\mathrm{aN}& =& {\mathrm{bn}}_1\mathrm{N}\\ & =& \mathrm{b}\left({\mathrm{n}}_1\mathrm{N}\right)\\ & =& \mathrm{bN}\\ & =& \mathrm{Na}\end{array}$

Thus, $\mathrm{aN}=\mathrm{Na}$, for every $\mathrm{a}\in \mathrm{N}$. This implies that$\mathrm{N}$ is a normal subgroup.

Hence, it is proved that$\mathrm{N}$ is a normal subgroup.