Question

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

Short answer

It is proved that$\mathrm{aK}=\mathrm{K}$ if and only if $\mathrm{a}\in \mathrm{K}$.

Step-by-step solution

Prove that  a∈K

Assume that$\mathrm{K}$ be a subgroup of a group$\mathrm{G}$ and consider $\mathrm{a}\in \mathrm{G}$. Now, we have to show that$\mathrm{aK}=\mathrm{K}$ if and only if $\mathrm{a}\in \mathrm{K}$.

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

$$\begin{gathered} \mathrm{ae}=\mathrm{a} \\ \in \mathrm{aK} \\ =\mathrm{K} \end{gathered}$$

It is proved that if$\mathrm{aK}=\mathrm{K}$ then $\mathrm{a}\in \mathrm{K}$.

Prove that aK=K

Assume that $\mathrm{a}\in \mathrm{K}$, then we have to prove that role="math" localid="1655799982044" $\mathrm{aK}=\mathrm{K}$.

Consider an element $\mathrm{x}\in \mathrm{aK}$, this implies that:

$\mathrm{x}=\mathrm{aK}$, for some$\mathrm{k}\in \mathrm{K}$

As both$\mathrm{a}$ and$\mathrm{k}$ belongs to$\mathrm{K}$ and$\mathrm{K}$ is a subgroup, this implies that$\mathrm{ak}\in \mathrm{K}$ which verify that role="math" localid="1655802908791" $\mathrm{aK}\subset \mathrm{K}$.

Now, verify that$\mathrm{K}\subset \mathrm{aK}$ as assume that $\mathrm{k}\in \mathrm{K}$. As$\mathrm{K}$ is a subgroup and $\mathrm{a}\in \mathrm{K}$, this implies that ${\mathrm{a}}^{-1}\mathrm{k}\in \mathrm{K}$. Therefore, we have:

$$\begin{gathered} \mathrm{k}={\mathrm{aa}}^{-1}\mathrm{k} \\ \in \mathrm{aK} \end{gathered}$$

This implies that$\mathrm{K}\subset \mathrm{aK}$.

Hence, $\mathrm{aK}=\mathrm{K}$.

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