Tag Archives: dihedral group

Compute in a group ring over Dih(8)

Let \alpha = r + r^2 - 2s and \beta = -3r^2 + rs be elements of the group ring \mathbb{Z}[D_8]. Compute the following: \beta\alpha, \alpha^2, \alpha\beta - \beta\alpha, and \beta\alpha\beta.


Evidently,

  1. \beta\alpha = -3 - 2r - 3r^3 + s + 6r^2s + r^3s
  2. \alpha^2 = 5 + r^2 + 2r^3 - 4r^2s - 4r^3s
  3. \alpha\beta - \beta\alpha = 2r - 2r^3 - s + r^2s
  4. \beta\alpha\beta = 15r + 10r^2 + 7r^3 - 21s - 6rs - 5r^2s

Free groups of rank at least 2 are nonabelian

Prove that if |S| > 1 then F(S) is nonabelian.


First we show that the free group of rank 2 is nonabelian. Suppose to the contrary that F(S) is abelian where |S| = 2; say S = \{a,b\}. Define \varphi : S \rightarrow D_{2n} by \varphi(a) = r, \varphi(b) = s. By the universal property of free groups, there exists a unique group homomorphism \Phi : F(S) \rightarrow D_{2n}. This homomorphism is clearly surjective, so that by the First Isomorphism Theorem, D_{2n} \cong F(S)/\mathsf{ker}\ \Phi. However, every quotient of an abelian group is abelian, but if n \geq 3, D_{2n} is not abelian. Thus F(S) is nonabelian.

Clearly every free group of rank at least 2 contains a subgroup which is free of rank 2. (For instance, use the universal property.) Thus no free group of rank at least 2 can be abelian.

Compute the higher centers of Dih(2ⁿ⁺¹)

Prove that for all n \geq 1 and 0 \leq k \leq n, Z_k(D_{2 \cdot 2^n}) = (D_{2 \cdot 2^n})^{n-k}.


We first consider the case n = 1. Now D_{2 \cdot 2^1} = D_4 \cong V_4 is abelian, so that Z_0(D_4) = 1 = [D_4,D_4] = D_4^1 and Z_1(D_4) = D_4 = D_4^0. Thus Z_k(D_4) = D_4^{1-k} for all 0 \leq k \leq 1.

Now let n \geq 2. Recall that D_{2 \cdot 2^n} is nilpotent of nilpotency class n, so that Z_n(D_{2 \cdot 2^n}) = D_{2 \cdot 2^n} = D_{2 \cdot 2^n}^0 and Z_0(D_{2 \cdot 2^n} = 1 = D_{2 \cdot 2^n}^n. We now consider the case 1 \leq k < n by induction on k; in particular, we compute Z_k(D_{2 \cdot 2^n}) = (D_{2 \cdot 2^n})^{n-k} = \langle r^{2^{n-k}} \rangle, by two separate induction arguments.

First we show that Z_k(D_{2 \cdot 2^n}) = \langle r^{2^{n-k}} \rangle by induction on k.

For the base case, note that since n \geq 2, by §2.2 #7, Z_1(D_{2 \cdot 2^n}) = \langle r^{2^{n-1}} \rangle.

For the inductive step, suppose Z_k(D_{2 \cdot 2^n}) = \langle r^{2^{n-1}} \rangle where 1 \leq k < k+1 < n. By definition, Z_{k+1}(D_{2 \cdot 2^n})/Z_k(D_{2 \cdot 2^n}) = Z(D_{2 \cdot 2^{n-k}}/Z_k(D_{2 \cdot 2^n})) = Z(D_{2 \cdot 2^{n-k}}/\langle r^{2^{n-k}} \rangle). Using the isomorphism defined in this previous exercise, and the calculation of the center in this previous exercise, we see that Z_{k+1}(D_{2 \cdot 2^n})/\langle r^{2^{n-k}} \rangle = \langle r^{2^{n-k-1}} \rangle / \langle r^{2^{n-k}} \rangle. Thus Z_{k+1}(D_{2 \cdot 2^n}) = \langle r^{2^{n-(k+1)}} \rangle.

Now we show that D_{2 \cdot 2^n}^{n-k} = \langle r^{2^{n-k}} \rangle for all 1 \leq k < n by induction on k.

For the base case, let k = n-1. Now D_{2 \cdot 2^n}^1 = [D_{2 \cdot 2^n}, D_{2 \cdot 2^n}]. Let x,y \in D_{2 \cdot 2^n} be arbitrary; there are 4 cases to consider.

  1. If x = r^a and y = r^b, then [x,y] = 1.
  2. If x = sr^a and y = r^b, then [x,y] = r^{2b}.
  3. If x = r^a and y = sr^b, then [x,y] = r^{2a}.
  4. If x = sr^a and y = sr^b, then [x,y] = r^{2(b-a)}.

Thus D_{2 \cdot 2^n}^1 = \langle r^2 \rangle = \langle r^{2^{n-k}} \rangle.

For the inductive step, suppose that for some 1 < k < n the conclusion holds, and consider k-1. We have D_{2 \cdot 2^n}^{n-(k-1)} = D_{2 \cdot 2^n}^{n-k+1} = [D_{2 \cdot 2^n},D_{2 \cdot 2^n}^{n-k}] = [D_{2 \cdot 2^n}, \langle r^{2^{n-k}} \rangle]. Let x \in D_{2 \cdot 2^n} and y \in \langle r^{2^{n-k}} \rangle; there are two cases to consider.

  1. If x = r^a and y = r^{b2^{n-k}}, then [x,y] = 1.
  2. If x = sr^a and y = r^{b2^{n-k}}, then [x,y] = r^{b2^{n-k+1}}.

Thus D_{2 \cdot 2^n}^{n-(k-1)} = \langle r^{2^{n-(k-1)}} \rangle.

In particular, for all n \geq 1 and 0 \leq k \leq n, we have Z_k(D_{2 \cdot 2^n}) = D_{2 \cdot 2^n}^{n-k} = \langle r^{2^{n-k}} \rangle.

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.

Construct Dih(8) as a semidirect product in two ways

Let p = 2 and check that the construction of the two nonabelian groups of order p^3 (in Example 7) is valid in this case. Prove that both of the resulting groups are isomorphic to D_8.


For the first construction, let H = Z_2 \times Z_2 and K = Z_2 = \langle x \rangle. Proposition 4.17(a) in the text does not apply for p=2 as it does for odd p; however, the comments after Proposition 4.17 demonstrate that for this special case, we still have \mathsf{Aut}(H) \cong GL_2(\mathbb{F}_2). Since (by Cauchy) \mathsf{Aut}(H) has an element \sigma of order 2, we can define \varphi : K \rightarrow \mathsf{Aut}(H) by x \mapsto \sigma. If H = \langle a \rangle \times \langle b \rangle, say \sigma(a,1) = (1,b) and \sigma(1,b) = (a,1). Then the semidirect product G_1 = H \rtimes_\varphi K has order 2^3 and is nonabelian since (a,x)(b,1) = (1,x) while (b,1)(a,x) = (ab,x).

For the second construction, let H = Z_4 and K = Z_2 = \langle x \rangle. Now since H is abelian, as in Example 1 we can define \psi : K \rightarrow \mathsf{Aut}(H) by \psi(x)(h) = h^{-1}. Then if G_2 = H \rtimes_\psi K, we have G_2 \cong D_8 a nonabelian group of order 8.

Now it remains to show that G_1 is isomorphic to D_8. To that end, it suffices to show that there exist elements r,s \in G_1 such that r^4 = s^2 = 1, s \notin \langle r \rangle, and rs = sr^{-1}.

Let r = (a,x) and s = (b,1). A straightforward calculation verifies that \langle r \rangle = \{ (a,x), (ab,1), (b,x), (1,1) \}, \langle (b,1) \rangle = \{ (b,1), (1,1) \}, and (a,x)(b,1) = (1,x) = (b,1)(b,x).

Dih(8n) is not isomorphic to the direct product of Dih(4n) and Cyc(2)

Prove that D_{8n} is not isomorphic to D_{4n} \times Z_2.


Suppose to the contrary that D_{8n} \cong D_{4n} \times Z_2. Then there exist subgroups H,K \leq D_{8n} such that (1) H and K are normal in D_{8n}, (2) H \cong D_{4n}, (3) K \cong Z_2, (4) H \cap K = 1, and (5) HK = D_{8n}.

Recall that D_{8n} has a unique cyclic subgroup of order 2n; namely \langle r^2 \rangle. It must be the case that \langle r^2 \rangle \leq H.

Now the subgroups of order 2 in D_{8n} are of the form \langle sr^i \rangle for 0 \leq i < 2n and \langle r^{2n} \rangle. Recall that if |x| = 2, then N_G(\langle x \rangle) = C_G(\langle x \rangle). In particular, because sr^i is noncentral in D_{8n}, \langle sr^i \rangle is not normal in D_{8n}. Thus K cannot be of this form. But then \langle r^{2n} \rangle \leq H, so that H \cap K \neq 1, and K cannot be of this form either. Thus we have a contradiction, and no such subgroups H and K exist.

Thus D_{8n} is not isomorphic to D_{4n} \times Z_2.

Exhibit six distinct nonabelian groups of order 16

Prove that D_{16}, Z_2 \times D_8, Z_2 \times Q_8, Z_4 \ast_\varphi D_8, QD_{16}, and M are nonisomorphic non-abelian groups of order 16. (Z_4 \ast_\varphi D_8 is described here and QD_{16} and M are described here and here, respectively.)


In §5.1 #6, we saw that all subgroups of Z_2 \times Q_8 are normal. Note that \langle s \rangle \leq D_{16} is not normal since rsr^{-1} = sr^6, so that Z_2 \times Q_8 \not\cong D_{16}; that \langle (x,s) \rangle \leq Z_2 \times D_8 is not normal since (1,r)(x,s)(1,r^{-1}) = (x,sr^2), so that Z_2 \times Q_8 \not\cong Z_2 \times D_8; and that \langle \tau \rangle \leq QD_{16} is not normal since \sigma \tau \sigma^{-1} = \tau \sigma^2, so that Z_2 \times Q_8 \not\cong QD_{16}.

The maximal subgroups of M are all abelian, while the maximal subgroups of QD_{16} and D_{16} are not. Thus M \not\cong QD_{16} and M \not\cong D_{16}. Similarly, 1 \times D_8 \leq Z_2 \times D_8 is a nonabelian maximal subgroup, so that M \not\cong Z_2 \times D_8, and 1 \times Q_8 \leq Z_2 \times Q_8 is a nonabelian maximal subgroup, so that M \not\cong Z_2 \times Q_8.

Note that 1 \times D_8, Z_2 \times \langle r \rangle, Z_2 \times \langle s,r^2 \rangle, and Z_2 \times \langle rs,r^2 \rangle are distinct maximal subgroups of Z_2 \times D_8, so that this group has at least 4 maximal subgroups. In particular, Z_2 \times D_8 \not\cong D_{16} and Z_2 \times D_8 \not\cong QD_{16}.

We previously computed the subgroup lattices of D_{16} and QD_{16}; D_{16} has 9 subgroups of order 2, while QD_{16} has 5. Thus D_{16} \not\cong QD_{16}.

The (relatively) hairy group to work with is Z_4 \ast_\varphi D_8.

Recall from §5.1 #13 that Z_4 \ast_\varphi D_8 has the presentation \langle a,b,c \ |\ a^4 = b^4 = c^2 = 1, a^2 = b^2, ab = ba, ac = ca, bc = cb^3 \rangle. We can see that every element of this group can be written in the form a^ic^jb^k, where 0 \leq i < 4, 0 \leq j < 2, and 0 \leq k < 2. This gives at most 16 "representations" of elements in Z_4 \ast_\varphi D_8, and we know by other means that this group has exactly 16 elements. Thus these representations are distinct.

In the following table, we compute the orders of the elements in Z_4 \ast_\varphi D_8.

x Reasoning |x|
1 1
a 4
a^2 2
a^3 4
c 2
ac |ac| = \mathsf{lcm}(|a|,|c|) 4
a^2c |a^2c| = \mathsf{lcm}(|a^2|,|c|) 2
a^3c |a^3c| = \mathsf{lcm}(|a^3|,|c|) 4
b 4
ab |ab| = \mathsf{lcm}(|a|,|b|) 4
a^2b |a^2b| = \mathsf{lcm}(|a^2|,|b|) 4
a^3b |a^3b| = \mathsf{lcm}(|a^3|,|b|) 4
cb cbcb = ccb^3b = 1 2
acb |acb| = \mathsf{lcm}(|a|,|cb|) 4
a^2cb |a^2cb| = \mathsf{lcm}(|a^2|,|cb|) 2
a^3cb |a^3cb| = \mathsf{lcm}(|a^3|,|cb|) 4

Note that Z_4 \ast_\varphi D_8 contains no elements of order greater than 4. Thus, since r \in D_{16} has order 8, Z_4 \ast_\varphi D_8 \not\cong D_{16}; since \sigma \in QD_{16} has order 8, Z_4 \ast_\varphi D_8 \not\cong QD_{16}; and since v \in M has order 8, Z_4 \ast_\varphi D_8 \not\cong M.

Note also that Z_4 \ast_\varphi D_8 contains 5 elements of order 2. Now (1,s), (1,sr), (1,sr^2), (1,sr^3), (x,1), and (x,s) are 6 distinct elements of order 2 in Z_2 \times D_8, so that Z_4 \ast_\varphi \not\cong Z_2 \times D_8.

Finally, note that (a,b) \in Z_2 \times Q_8 has order 2 only if |b| \in \{1,2\}. Thus the only elements of order 2 in Z_2 \times Q_8 are (x,1), (1,-1), and (x,-1). In particular, Z_4 \ast_\varphi D_8 \not\cong Z_2 \times Q_8.

Thus these groups are nonisomorphic nonabelian groups of order 16.

Compute presentations for a given central product of groups

Give presentations for the groups Z_4 \ast_\varphi D_8 and Z_4 \ast_\psi Q_8 constructed in a previous example.


Note that Z_4 \times D_8 is generated by (x,1), (1,r), and (1,s). Hence Z_4 \ast_\varphi D_8 is generated by the images of these elements under the natural projection; moreover, the relations satisfied by these also hold in the quotient. We have an additional relation: note that (x,1)^2(1,r)^2 = (x^2,r^2) \in Z. Thus Z_4 \ast_\varphi D_8 = \langle a,b,c \ |\ a^4 = b^4 = c^2 = 1, a^2 = b^2, ab = ba, ac = ca, bc = cb^3 \rangle.

Similarly, Z_4 \ast_\psi Q_8 is generated by the natural images of (x,1), (1,i), and (1,j), and we have an additional relation because (x,1)^2(1,i)^2 = (x^2,-1) \in Z. Thus Z_4 \ast_\psi Q_8 = \langle a,b,c \ |\ a^4 = b^4 = c^4 = 1, a^2 = b^2 = c^2, ab = ba, ac = ca, bc = cb^3 \rangle

Basic properties of the central product of groups

Let A and B be groups. Suppose Z_1 \leq Z(A) and Z_2 \leq Z(B) are subgroups and that there exists an isomorphism \varphi : Z_1 \rightarrow Z_2. Define Z \leq A \times B by Z = \{ (x, \varphi(x)^{-1}) \ |\ x \in Z_1 \}. Note that Z is normal in A \times B since Z \leq Z(A \times B), using a previous theorem. We denote by A \ast_\varphi B the quotient (A \times B)/ Z. (In particular, A \ast_\varphi B depends on \varphi.) Think of A \ast_\varphi B as the direct product A \times B “collapsed” by identifying each element x \in Z_1 with its \varphi-image in Z_2.

  • Prove that the images of A and B in A \ast_\varphi B are isomorphic to A and B, respectively, and that these images intersect in a central subgroup isomorphic to Z_1. Find |A \ast_\varphi B|.
  • Let Z_4 = \langle x \rangle. Let D_8 = \langle r,s \rangle and Q_8 = \langle i,j \rangle as usual. Let Z_4 \ast_\varphi D_8 be the central product of Z_4 and D_8 which identifies x_2 and r^2. (I.e. Z_1 = \langle x^2 \rangle, Z_2 = \langle r^2 \rangle, and \varphi(x^2) = r^2.) Let Z_4 \ast_\psi Q_8 be the central product of Z_4 and Q_8 which identifies x^2 and -1. (I.e. Z_1 = \langle x^2, Z_2 = \langle -1 \rangle, and \psi(x^2) = -1.) Prove that Z_4 \ast_\varphi D_8 \cong Z_4 \ast_\psi Q_8.

  1. Let \pi : A \times B \rightarrow (A \times B)/Z denote the canonical projection, and identify A with A \times 1 and B with 1 \times B in A \times B.

    Let (a_1,1), (a_2,1) \in A \times 1, and suppose \pi((a_1,1)) = \pi((a_2,1)). Then (a_2a_1^{-1},1) \in Z, so that a_2a_1^{-1} = \pi(1) = 1. Thus a_1 = a_2, and hence \pi|_A is injective. Similarly, \pi|_B is injective.

    Note that the restriction \pi|_{Z_1} is also injective. Suppose (x,y)Z \in \pi[A] \cap \pi[B]; then for some a \in A and b \in B we have (x,y)Z = (a,1)Z = (1,b)Z. Thus (a,b^{-1}) \in Z; by definition, then, a \in Z_1 and b = \varphi(a). Note that for all (z,w)Z \in A \ast_\varphi B we have (z,w)Z(a,1)Z = (za,w)Z = (az,w)Z (a,1)Z(z,w)Z, so that \pi[A] \cap \pi[B] is in the center of A \ast_\varphi B. Moreover, (x,y)Z \in \mathsf{im}\ \pi|_{Z_1}. Conversely, if (z,1) ]in Z_1 \times 1, we have (z,1)Z = (z,1)(z^{-1}, \varphi(z))Z = (1,\varphi(z))Z, so that \mathsf{im}\ \pi|_{Z_1} \subseteq \pi[A] \cap \pi[B]. Then \mathsf{im}\ \pi|_{Z_1} = \pi[A] \cap \pi[B], and by the First Isomorphism Theorem, Z_1 \cong \pi[A] \cap \pi[B]. In addition, \pi[A] \cap \pi[B] is a central subgroup of A \ast_\varphi B.

    Finally, by Lagrange write |A| = n|Z_1| and |B| = m|Z_2|. Note that |Z_1| = |Z_2|. Now |A \ast_\varphi B| = |A \times B|/|Z| = nm|Z_1|^2/|Z_1| = nm|Z_1|. We may also write this equation in the form |A \ast_\varphi B| = |A| \cdot [B : Z_2] = |B| \cdot [A : Z_1].

  2. Define \overline{\alpha} : \{(x,1), (1,i), (1,j)\} \rightarrow Z_4 \times D_8 as follows: \overline{\alpha}((x,1)) = (x,1), \overline{\alpha}((1,i)) = (1,r), and \overline{\alpha}((1,j)) = (x,s). Because \{(x,1),(1,i),(1,j)\} generates Z_4 \times Q_8 and these images satisfy the relations \overline{\alpha}((x,1))^4 = \overline{\alpha}((1,i))^4 = \overline{\alpha}((1,j))^4 = 1, \overline{\alpha}((x,1))\overline{\alpha}((1,i)) = \overline{\alpha}((1,i))\overline{\alpha}((x,1)), \overline{\alpha}((x,1))\overline{\alpha}((1,j)) = \overline{\alpha}((1,j))\overline{\alpha}((x,1)), and \overline{\alpha}((1,i))\overline{\alpha}((1,j)) = \overline{\alpha}((1,j))\overline{\alpha}((1,i))^3, \overline{\alpha} extends to a homomorphism \alpha : Z_4 \times Q_8 \rightarrow Z_4 \times D_8.

    If we let \pi denote the natural projection, \pi \circ \alpha is a mapping Z_4 \times Q_8 \rightarrow Z_4 \ast_\varphi D_8. We claim that Z \leq Z_4 \times Q_8 is contained in the kernel of this mapping; it suffices to show this for the (unique) nonidentity element of Z, namely (x^2,-1). In fact, (\pi \circ \alpha)(x^2,-1) = \pi(\alpha(x^2, i^2)) = \pi(\alpha((x,1)^2(1,i)^2) = \pi(x^2,r^2) = 1. Thus Z \leq \mathsf{ker}(\pi \circ \alpha). By the remarks on page 100 in the text, there exists a unique group homomorphism \theta : Z_4 \ast_\psi Q_8 \rightarrow Z_4 \ast_\varphi D_8 such that \theta \circ \pi = \pi \circ \alpha, and in particular, \theta(tZ) = (\pi \circ \alpha)(t).

    We now show that \mathsf{ker}\ \theta = 1. Suppose \theta((t,u)Z) = 1. Then \pi(\alpha(t,u)) = 1; note that (t,u) = (x^a,i^bj^c) for some integers a,b,c, and that \alpha(t,u) = (x^{a+c},r^bs^c). Then (x^{a+c},r^bs^c) \in \{ (1,1), (x^2,r^2) \}.

    If (x^{a+c},r^bs^c) = (1,1), then c \equiv 0 mod 2 and b \equiv 0 mod 4. Now if c \equiv 0 mod 4, then a \equiv 0 mod 4 and (t,u) = (1,1) \in Z. If c \not\equiv 0 mod 4, then a \equiv 0 mod 2 and (t,u) = (x^2,-1) \in Z.

    If (x^{a+c},r^bs^c) = (x^2,r^2), then c \equiv 0 mod 2 and b \equiv 2 mod 4. If c \equiv 0 mod 4, then a \equiv 2 mod 4 and (t,u) = (x^2,-1) \in Z. If c \not\equiv 0 mod 4, then a \equiv 0 mod 4 and (t,u) = (1,1) \in Z.

    Thus \mathsf{ker}\ \theta = 1, hence \theta is injective.

    Now by part (a), we see that both Z_4 \ast_\varphi D_8 and Z_4 \ast_\psi Q_8 have order 16. Thus \theta is an isomorphism.

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}.