Let H = ⟨R n ⟩. Show that H is normal in D2n.

Question

Let n ∈ Z≥2, and let D2n be the symmetry group of a regular 2n-gon. Recall that D2n = {I, R, R^2, . . . , R^2n−1, F, RF, R^2F, . . . , R^(2n−1)F} and F R = R^(2n−1)F. Let H = ⟨Rn⟩. Show that H is normal in D2n.

Work: I know if I just plug in values its normal but I don't know how to show it arbitrarily in a proof.

@DietrichBurde, this question is about the dihedral group of order $4n$. — lhf, Mar 16 '18 at 16:25

lhf · Answer 1 · 2018-03-16T16:18:09.127

0

Hint: Assuming you mean $H = \langle R^n \rangle$, prove that $FR^nF^{-1}=R^n$, using $FRF^{-1}=R^{-1}$.

edited Mar 16 '18 at 16:18

answered Mar 16 '18 at 16:12

lhf

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Let H = ⟨R n ⟩. Show that H is normal in D2n.

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