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

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  • $\begingroup$ Rotating by 45 degree twice does not get you back where you started. $\endgroup$ Commented Sep 24, 2015 at 8:40
  • $\begingroup$ I think the $r_2$ in your 'Cayley table' should be $r_0$? $\endgroup$
    – Servaes
    Commented Sep 24, 2015 at 12:59
  • $\begingroup$ It was a typo. I replaced $r_2$ with $r_0$ in the first column. @TobiasKildetoft $\endgroup$ Commented Sep 24, 2015 at 16:58

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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?

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  • $\begingroup$ So, from here I just need to show that $\{ r_0, r_4, s_0, s_4 \}$ is a subgroup of $D_8$, i.e., show that this subset is closed under composition, associative, a unique identity exists, and that each element has an inverse? @Servaes $\endgroup$ Commented Sep 24, 2015 at 17:03
  • $\begingroup$ Indeed. Note that associativity follows from the fact that $D_8$ is itself a group; you never need to check this for a subgroup. $\endgroup$
    – Servaes
    Commented Sep 24, 2015 at 17:17
  • $\begingroup$ Oh yeah, associativity is inherited from the group $D_8$. Thank you very much for your help! $\endgroup$ Commented Sep 24, 2015 at 17:18

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