Determine the subgroup of $D_8$ generated by $r_4$ and $s_0$.

Question

I am wondering if my work for computing $\langle r_4 , s_0 \rangle$ in $D_8$ is correct.

Here $r$ denotes rotation of 45 degrees and $s$ denotes reflections about the lines of symmetries.

$D_8 = \{ r_0, r_1, r_2, r_3, r_4, r_5, r_6, r_7, s_0, s_1, s_2, s_3, s_4, s_5, s_6, s_7 \}$

Then $\langle r_4 \rangle = \{r_0, r_4 \}$ since $r_4 \circ r_4 = r_0$ and $\langle s_0 \rangle = \{ r_0, s_0 \}$ since $s_0 \circ s_0 = r_0$.

Then I use the following Caley table to compute $\langle r_4, s_0 \rangle.$

\begin{array}{c|cc} \circ & r_0 & r_4 \\ \hline r_0 & r_0 & r_4 \\ s_0 & s_0 & s_4 \end{array}

Thus, $\langle r_4, s_0 \rangle = \{ r_0, r_4, s_0, s_4 \}$

Answer 1

You cannot use a Cayley table in this way to 'compute' $\langle r_4,s_0\rangle$. You have multiplied a few elements from $\{r_0,r_4,s_0,s_4\}$ together, but what ensures you that you have all elements of $\langle r_4,s_0\rangle$?

Here's another approach: The subgroup $\langle r_4,s_0\rangle$ of $D_8$ is the smallest subgroup of $D_8$ containing both $r_4$ and $s_0$. Your calculation already show that $$\{r_0,r_4,s_0,s_4\}\subset\langle r_4,s_0\rangle.$$ If you can show that $\{r_0,r_4,s_0,s_4\}$ is a subgroup of $D_8$ then it follows that $$\langle r_4,s_0\rangle=\{r_0,r_4,s_0,s_4\}.$$ Can you take it from here?