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

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

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

## 1 Answer

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?

• 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 Commented Sep 24, 2015 at 17:03
• Indeed. Note that associativity follows from the fact that $D_8$ is itself a group; you never need to check this for a subgroup. Commented Sep 24, 2015 at 17:17
• Oh yeah, associativity is inherited from the group $D_8$. Thank you very much for your help! Commented Sep 24, 2015 at 17:18