## 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 $D_{2n}$.