How to prove that the center of the dihedral group $D_{2n}$ is $\{1,r^{n}\}$ and the center of $D_{2n-1}$ is $\{1\}$?

I don't know how to prove it in this general case.


closed as off-topic by Derek Holt, Shailesh, user296602, user91500, colormegone Jan 27 '16 at 4:36

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  • $\begingroup$ Does this help ? $\endgroup$ – Watson Jan 11 '16 at 21:28
  • 2
    $\begingroup$ It's really awkward to structure the title so that "$D_{2n} = \{1, r^n\}"$ is a phrase. I understand you're saying the center, and not the whole group, is what the equality refers to, but it just looks so wrong. $\endgroup$ – pjs36 Jan 11 '16 at 21:56

Let us conjugate a general element of the form $r^as$ by another element $r^bs$. $$(r^bs)r^as(sr^{-b})=r^bsr^ar^{-b}=r^{2b-a}s$$ Since in general this depends on $b$, an element of this form cannot be in the center. Now let's conjugate $r^a$ by $r^bs$: $$(r^bs)r^a(sr^{-b})=r^{-a}$$ For completeness we note $$r^br^ar^{-b}=r^a$$ Thus the only way an element $r^a$ can be in the center is if $r^a=r^{-a}$, meaning the rotation commutes with reflections. This means $$r^{2a}=1$$ $a=0$ will always work; for a dihedral group of order $2(2n)$, we can also take $a=n$. For a dihedral group of order $2(2n-1)$ there is no $a$ such that $r^a$ has order $2$, hence the center is trivial.

  • $\begingroup$ @bob no problem. $\endgroup$ – Matt Samuel Jan 12 '16 at 12:12

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