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I am looking for the order of the stabilizer group of $D_{10}$. I know that ${G_x} = \{g \in G : gx = x\}$. I am curious what to use for $x$ though? Should I just cycle through elements of $D_{10}$ and see how many $x$'s are fixed by each element of $D_{10}$?

My approach so far (trying the above) is:

For identity element, 10 vertices are all fixed.

For any rotation, no vertices are fixed.

For any reflection, similarly, no vertices are fixed (draw out a 10-gon to see this).

Is this right? I know that the order of the stabilizer of $D_{10}$ is not 10, so I'm sure I made a mistake somewhere. Did I not consider enough points?

Thanks.

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    $\begingroup$ For half of the reflections, two vertices are fixed. $\endgroup$ – Arthur Nov 30 '14 at 22:02
  • $\begingroup$ @Arthur Could you explain how? I don't really see it. $\endgroup$ – jstnchng Nov 30 '14 at 22:03
  • $\begingroup$ Say that you reflect any two vertices. Wouldn't you have the other 8 moving as well? $\endgroup$ – jstnchng Nov 30 '14 at 22:04
  • $\begingroup$ My mistake. I see it now, thank you. $\endgroup$ – jstnchng Nov 30 '14 at 22:14
  • $\begingroup$ @Arthur is it true that there are only 20 symmetries then? I'm fairly sure the number of symmetries is 160. $\endgroup$ – jstnchng Nov 30 '14 at 22:20

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