Dih(2n) is nilpotent if and only if n is a power of 2

Prove that D_{2n} is nilpotent if and only if n is a power of 2.


(\Rightarrow) Suppose D_{2n} is nilpotent. Let p be an odd prime dividing n. Then r^{n/p} is an element of order p in D_{2n}; in particular, r^{n/p} \neq r^{-n/p}. Now |s| = 2 and |r^{n/p}| = p are relatively prime, so that, by §6.1 #9, sr^{n/p} = r^{n/p}s; a contradiction. Thus no odd primes divide n, and we have n = 2^k.

(\Leftarrow) We proceed by induction on k, where n = 2^k.

For the base case, k = 0, note that D_{2 \cdot 2^0} \cong Z_2 is abelian, hence nilpotent.

For the inductive step, suppose D_{2 \cdot 2^k} is nilpotent. Consider D_{2 \cdot 2^{k+1}}; we have Z(D_{2 \cdot 2^{k+1}}) = \langle r^{2^k} \rangle, and by §3.1 #34, D_{2 \cdot 2^{k+1}}/Z(D_{2 \cdot 2^{k+1}}) \cong D_{2 \cdot 2^k} is nilpotent. By §6.1 #6, D_{2 \cdot 2^{k+1}} is nilpotent.

Moreover, we show that for k \geq 1, D_{2\cdot 2^k} is nilpotent of nilpotence class k, by induction on k.

For the base case, k = 1, D_{2\cdot 2} \cong V_4 is abelian, and thus of nilpotence class 1.

Suppose now that D_{2 \cdot 2^k} is nilpotent of nilpotence class k. Since D_{2 \cdot 2^k} \cong D_{2 \cdot 2^{k+1}}/Z(D_{2 \cdot 2^{k+1}}), by a previous lemma, D_{2 \cdot 2^{k+1}} is nilpotent of nilpotence class k+1.

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