How can we show by a direct group-theoretic proof that the dihedral group $D_{2n}$ is a Frobenius group iff $n$ is an odd number?
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Hint: My assertion is that $D_{2n}$ acts on the vertices of respective $n$-gon. Non-trivial rotations have no fixed points. Further: 1) If $n$ is odd then each reflection has precisely one fixed point. So, $D_{2n}$ is Frobenius. 2) If $n$ is even then every reflection has precisely two fixed points. So, $D_{2n}$ is not Frobenius.
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1$\begingroup$ For the $n$ even case, this shows only that $D_{2n}$ is not Frobenius in its action as the symmetry group of an $n$-gon. For a complete answer you would need to show that it has no (faithful) action as a Frobenius group. $\endgroup$ – Derek Holt Aug 11 '14 at 17:11
