Question

Let $\mathit{K}$be a subgroup of a group$\mathit{G}$and let$\mathit{a}\mathbf{\in }\mathit{G}$Prove that$\mathit{a}\mathit{K}\mathbf{=}\mathit{K}$if and only if$\mathit{a}\mathbf{\in }\mathit{K}$

Short answer

It is proved that$\mathit{a}\mathit{K}\mathbf{=}\mathit{K}$if and only iflocalid="1657361007298" role="math" $\mathit{a}\mathbf{\in }\mathit{K}\mathbf{.}$

Step-by-step solution

Prove that a∈K

Assume that $\mathit{K}$be a subgroup of a grouprole="math" localid="1657360929007" $\mathit{G}$and consider $\mathit{a}\mathbf{\in }\mathit{G}\mathbf{.}$Now, we have to show that$\mathit{a}\mathit{K}\mathbf{=}\mathit{K}$if and only if $\mathit{a}\mathbf{\in }\mathit{K}\mathbf{.}$

As $\mathit{K}\mathbf{=}\mathit{K}$, since $\mathit{K}$is a subgroup with the identity$\mathit{e}\mathbf{\in }\mathit{K}$. This implies that:

$$\begin{gathered} \mathit{a}\mathit{e}\mathbf{=}\mathit{a} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\in }\mathit{a}\mathit{K} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{K}\mathbf{} \end{gathered}$$

It is proved that if $\mathit{a}\mathit{K}\mathbf{=}\mathit{K}$then$\mathit{a}\mathbf{\in }\mathit{K}$.

Prove that aK=K

Assume that $\mathit{a}\mathbf{\in }\mathit{K}\mathbf{,}$then we have to prove that $\mathit{a}\mathit{K}\mathbf{=}\mathit{K}\mathbf{.}$

Consider an element $\mathit{x}\mathbf{\in }\mathit{a}\mathit{K}\mathbf{,}$this implies that:

localid="1657360990870" $\mathit{x}\mathbf{=}\mathit{a}\mathit{k}\mathbf{,}\mathbf{\hspace{0.33em}}$for some localid="1657379734729" $\mathit{k}\mathbf{\in }\mathit{K}$

As both $\mathit{a}$and $\mathit{b}$belongs to $\mathit{K}$and $\mathit{K}$is a subgroup, this implies that $\mathit{a}\mathit{k}\mathbf{\in }\mathit{K}\mathbf{}$

which verify that$\mathit{a}\mathit{K}\mathbf{\subset }\mathit{K}\mathbf{.}$

Now, verify that $\mathit{K}\mathbf{\subset }\mathit{a}\mathit{K}$as assume that $\mathit{k}\mathbf{\in }\mathit{K}\mathbf{.}$As $\mathit{K}$is a subgroup and $\mathit{a}\mathbf{\in }\mathit{K}\mathbf{,}\mathbf{}$

this implies that ${\mathit{a}}^{\mathbf{(}}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{k}\mathbf{\in }\mathit{K}\mathbf{.}$Therefore, we have:

$$\begin{gathered} \mathit{k}\mathbf{=}\mathit{a}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{k} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{\in }\mathit{a}\mathit{k} \end{gathered}$$

This implies that $\mathit{K}\mathbf{\subset }\mathit{a}\mathit{K}\mathbf{.}$

Hence,$\mathit{a}\mathit{K}\mathbf{=}\mathit{K}\mathbf{.}$

Hence, it is proved that$\mathit{a}\mathit{K}\mathbf{=}\mathit{K}$if and only if$\mathit{a}\mathbf{\in }\mathit{K}\mathbf{.}$