Computation of the number of Sylow 2-subgroups in Dih(2n)

Let 2n = 2^ak where k is odd. Prove that the number of Sylow 2-subgroups of D_{2n} is k.

We begin with several lemmas.

Lemma 1: If n = km, then D_{2m} \leq D_{2n}. Proof: The homomorphism defined on the generators of D_{2m} by s \mapsto s and r \mapsto r^k is injective. \square

Lemma 2: Let G be a finite group and p a prime. If H \leq G is a subgroup such that p does not divide [G:H], then every Sylow p-subgroup of H is a Sylow p-subgroup of G. Proof: We have |G| = p^am and |H| = p^bn, with p not dividing m and n and n|m; since p does not divide the index of H in G, a=b. Hence any Sylow p subgroup of H has order p^a, and thus is a Sylow p-subgroup in G. \square

Now to the main result.

Suppose n is odd. Then k = n and every Sylow 2-subgroup has order 2. In this case, the Sylow 2-subgroups correspond to the elements of order 2 in D_{2n}. There are exactly n of these: elements of the form sr^b for 0 \leq b < n. Thus n_2 = k.

Suppose now that n = 2 \ell is even; then a \geq 2. We proceed by induction on the width of k. (Recall that the width of an integer is the number of prime factors which divide it, including multiplicity.) Let P be a Sylow 2-subgroup of D_{2n}.

For the base case, if the width of k is 0, we have k = 1. Then P = D_{2n}. In this case, P is the unique Sylow 2-subgroup, so that n_2 = 1.

For the inductive step, suppose that for any integer k of width at most t, the number of Sylow 2-subgroups of D_{2^ak} is k. Let k be an integer of width t+1. Then k = qm for some prime q and some width t integer m; without loss of generality, we may assume that q is the smallest prime divisor of n, so that q \leq m. By the induction hypothesis, D_{2^am} has m distinct Sylow 2-subgroups. By Lemma 1, D_{2^am} \leq D_{2^aqm}. Note also that 2 does not divide [D_{2^aqm} : D_{2^am}] = q, so that by Lemma 2, D_{2^am} has at least m distinct Sylow 2-subgroups.

Now by Sylow’s Theorem, n_2 divides k = qm. Thus either n_2 = m or n_2 = k. Note that there exist elements of order 2 in D_{2n} which are not in (the image of) D_{2m}; for instance, sr, since otherwise we have (sr^m)(sr)(r^m) = r in the image of D_{2m} in D_{2n}, a contradiction. By Sylow’s Theorem, every 2-subgroup (For instance \langle sr \rangle) is contained in some Sylow 2-subgroup. Thus there must exist at least one other Sylow 2-subgroup in D_{2n}; hence n_p = k.

By induction, the result is proved for all D_{2n}.

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