1

What are the conjugacy classes for the dihedral group $D_8$ of order 16?

What are its subgroups of order $4$, and which of them are normal subgroups?

I know that $\{e\} ,\{r^2,r^6\},\{r,r^7\},\{r^3,r^5\},\{r^4\},\{s,r^2s,r^4s,r^6s\}, \{rs,r^3s,r^5s,r^7s\}$ are the conjugacy classes, but I don't understand how they derived these in my book; they just mentioned it without explanation.

Jonas Meyer
  • 55,715
M Alrantisi
  • 85
  • 5
  • I bet you could at least do some work on this...otherwise one could suspect you want someone else to do all the work for you. Please do show your own work on this. – Timbuc Dec 13 '14 at 11:01
  • I know that {e} ,{r^2,r^6},{r,r^7},{r^3,r^5},{r^4},{so,r^s2,r^4so,r^6s},{rs,r^s3,r^5,r^6s} are conjuicy class but I don't know how to derive them as r rotation and a reflection – M Alrantisi Dec 13 '14 at 12:36
  • Why would you need to do such a thing?? – Timbuc Dec 13 '14 at 13:13
  • I would like to try every single step as I don't understand how they derived these conjugate classes in my book they just mentioned without explanation – M Alrantisi Dec 13 '14 at 13:22
  • @M alrantisi A first possible try could be to separate elements according to their orders, as conjugate elements have the same order. This doesn't seem hard at all to do with a group of order $;16;$ , though it can be a little time consuming. – Timbuc Dec 13 '14 at 13:24
  • OK thanks but what about subgroups of order 4 I can't have any of them – M Alrantisi Dec 13 '14 at 13:39
  • @M $$D_{16}=\langle; r,s;:;;r^2=s^8=1;,;;rsr=r^{7};\rangle$$So you can take for example$$N:=\langle;s^2;\rangle;,;;M:=\langle;s^6;\rangle$$ which are two subgroups of order four. Can you find more? – Timbuc Dec 13 '14 at 13:43
  • It's wrong presentation here as rsr=r^7 – M Alrantisi Dec 13 '14 at 13:44
  • @M Of course: that was a mistake of mine as I got confused with $;D_{32};$ . It must be $;7;$ there. – Timbuc Dec 13 '14 at 13:45
  • So can you guide me through finding subgroups – M Alrantisi Dec 13 '14 at 13:54
  • Why not first do what I proposed in my comment above: separate first the elements of $;G;$ according to their order. It's not problem here since the only possible orders here are $;1,2,4,8,16;$ . The first one is trivial, and the last one is impossible as $;G;$ isn't even abelian, much less cyclic. – Timbuc Dec 13 '14 at 14:09
  • I got a normal subgroup {e,r^2,r^4,r^6} – M Alrantisi Dec 13 '14 at 14:17
  • possible duplicate of Conjugate class in the dihedral group. See also http://math.stackexchange.com/questions/1064755/questions-about-the-dihedral-group-d-8. – Dietrich Burde Dec 13 '14 at 19:25

0 Answers0