Question

Let $\mathit{K}$be a subgroup localid="1657364298487" $\mathbf{\{}{\mathit{r}}_{\mathbf{0}}\mathbf{,}\mathit{v}\mathbf{\}}$of localid="1657364318199" ${\mathit{D}}_{\mathbf{4}}$. Show that localid="1657364309859" ${\mathit{r}}_{\mathbf{1}}\mathbf{\equiv }\mathit{t}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$andlocalid="1657361926143" ${\mathit{r}}_{\mathbf{2}}\mathbf{\equiv }\mathit{h}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$but ${\mathit{r}}_{\mathbf{1}}\mathbf{\circ }{\mathit{r}}_{\mathbf{2}}\mathbf{}\mathbf{\not\equiv }\mathit{t}\mathbf{\circ }\mathit{h}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$.

Short answer

It is proved that ${\mathit{r}}_{\mathbf{1}}\mathbf{\equiv }\mathit{t}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$,and${\mathit{r}}_{\mathbf{2}}\mathbf{\equiv }\mathit{h}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$but${\mathit{r}}_{\mathbf{1}}\mathbf{\circ }{\mathit{r}}_{\mathbf{2}}\mathbf{\not\equiv }\mathit{t}\mathbf{\circ }\mathit{h}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$.

Step-by-step solution

Prove that r1≡t(modK)and r2≡h(modK)

Assume that $\mathit{K}$be the subgroup of${\mathit{D}}_{\mathbf{4}}$. We have to show that ${\mathit{r}}_{\mathbf{1}}\mathbf{\equiv }\mathit{t}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$and${\mathit{r}}_{\mathbf{2}}\mathbf{\equiv }\mathit{h}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$.

Since we have:

role="math" localid="1657364067841" $$\begin{gathered} {\mathit{r}}_{\mathbf{-}\mathbf{1}}{\mathit{t}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{=}{\mathit{r}}_{\mathbf{1}}\mathit{t} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{V} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\in }\mathit{K} \end{gathered}$$

role="math" localid="1657364078822" $$\begin{gathered} {\mathit{r}}_{\mathbf{2}}{\mathit{h}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{=}{\mathit{r}}_{\mathbf{2}}\mathit{h} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{V} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\in }\mathit{K} \end{gathered}$$

This implies that ${\mathit{r}}_{\mathbf{1}}\mathbf{\equiv }\mathit{t}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$,and${\mathit{r}}_{\mathbf{2}}\mathbf{\equiv }\mathit{h}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}\mathbf{.}$Hence, it is proved that

${\mathit{r}}_{\mathbf{1}}\mathbf{\equiv }\mathit{t}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$, and${\mathit{r}}_{\mathbf{2}}\mathbf{\equiv }\mathit{h}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$.

Prove that r1∘r2≠t∘h(modK)

It is observed that ${\mathit{r}}_{\mathbf{1}}\mathbf{\circ }{\mathit{r}}_{\mathbf{2}}\mathbf{=}{\mathit{r}}_{\mathbf{3}}$and$\mathit{t}\mathbf{\circ }\mathit{h}\mathbf{=}{\mathit{r}}_{\mathbf{1}}$.As we know that ${\mathit{r}}_{\mathbf{1}}\mathbf{\ne }{\mathit{r}}_{\mathbf{3}}$.This implies that${\mathit{r}}_{\mathbf{1}}\mathbf{\circ }{\mathit{r}}_{\mathbf{2}}\mathbf{\not\equiv }\mathit{t}\mathbf{\circ }\mathit{h}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}\mathbf{.}$

Hence, it is proved thatrole="math" localid="1657363782916" ${\mathit{r}}_{\mathbf{1}}\mathbf{\equiv }\mathit{t}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$,and${\mathit{r}}_{\mathbf{2}}\mathbf{\equiv }\mathit{h}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$but${\mathit{r}}_{\mathbf{1}}\mathbf{\circ }{\mathit{r}}_{\mathbf{2}}\mathbf{\not\equiv }\mathit{t}\mathbf{\circ }\mathit{h}\mathbf{\left(}\mathit{m}\mathit{o}\mathit{d}\mathit{K}\mathbf{\right)}$.