## Tag Archives: group homomorphism

### A class of group homomorphisms from CC to GL(n,CC)

Fix an $n \times n$ matrix $M$ over $\mathbb{C}$. Define a mapping $\psi_M : \mathbb{C} \rightarrow \mathsf{GL}_n(\mathbb{C})$ by $\alpha \mapsto \mathsf{exp}(M\alpha)$. Prove that $\psi_M$ is a group homomorphism on $(\mathbb{C},+)$.

Note that $\mathsf{exp}(M\alpha)$ is nonsingular by this previous exercise, so that $\psi_M$ is properly defined.

Now $\psi_M(\alpha + \beta) = \mathsf{exp}(M(\alpha+\beta))$ $= \mathsf{exp}(M\alpha + M\beta)$ $= \mathsf{exp}(M\alpha) \cdot \mathsf{exp}(M\beta)$ by this previous exercise, which equals $\psi_M(\alpha) \psi_M(\beta)$. So $\psi_M$ is a group homomorphism.

### If A² is principal over an algebraic number field, then A² is principal over any extension

Let $K_1$ and $K_2$ be algebraic number fields with $K_1 \subseteq K_2$, and having integer rings $\mathcal{O}_1$ and $\mathcal{O}_2$, respectively. Suppose $(A)_{\mathcal{O}_1} \subseteq \mathcal{O}_1$ is an ideal such that $(A)^2$ is principal. Show that $(A)_{\mathcal{O}_2}^2$ is principal in $\mathcal{O}_2$.

We will let $G_1$ and $G_2$ denote the ideal class groups of $K_1$ and $K_2$, respectively. For brevity, if $S \subseteq \mathcal{O}_1$ is a subset, then we will denote by $(S)_1$ the ideal generated by $S$ in $\mathcal{O}_1$ and by $(S)_2$ the ideal generated by $S$ in $\mathcal{O}_2$.

We claim that if $(A)_1 \sim (B)_1$, then $(A)_2 \sim (B)_2$. To see this, suppose we have $\alpha$ and $\beta$ such that $(\alpha)_1(A)_1 = (\beta)_1(B)_1$. Then $(\alpha A)_1 = (\beta B)_1$. Now if $x = \sum r_i \alpha a_i \in (\alpha A)_2$, then each $\alpha a_i$ is in $(\beta B)_1$, and so has the form $\sum s_{i,j} \beta b_j$. So $x \in (\beta B)_2$. Conversely, $(\beta B)_2 \subseteq (\alpha A)_2$, and so $(A)_2 \sim (B)_2$.

The mapping $A \mapsto (A)_2$ then induces a well-defined mapping $\Psi : G_1 \rightarrow G_2$ given by $[A] \mapsto [(A)_2]$. Moreover, since $\Psi([A][B]) = \Psi([AB])$ $= [(AB)_2]$ $= [(A)_2][(B)_2]$ $= \Psi(A) \Psi(B)$, $\Psi$ is a group homomorphism.

In particular, if $A^2 \sim (1)$ in $\mathcal{O}_1$, then $\Psi(A) \sim (1)$ in $\mathcal{O}_2$, as desired.

### Some properties of a commutative diagram of groups with exact rows

Consider the following commutative diagram of groups.

A commutative diagram of groups

Suppose the rows are exact. That is, $\mathsf{ker}\ \varphi_k = \mathsf{im}\ \psi_k$ and $\mathsf{ker}\ \chi_k = \mathsf{im}\ \varphi_k$ for $k \in \{1,2\}$. Prove the following.

1. If $\alpha$ is surjective and $\beta$ and $\delta$ are injective, then $\gamma$ is injective.
2. If $\delta$ is injective and $\alpha$ and $\gamma$ are surjective, then $\beta$ is surjective.

1. Let $c \in \mathsf{ker}\ \gamma$. Note that $\chi_2(\gamma(c)) = 1$, so that $1 = (\chi_2 \circ \gamma)(c) = (\delta \circ \chi_1)(c) = \delta(\chi_1(c))$. Since $\delta$ is injective, $\chi_1(c) = 1$. Thus $c \in \mathsf{ker}\ \chi_1$, and so $c \in \mathsf{im}\ \varphi_1$. Say $b \in B_1$ such that $c = \varphi_1(b)$. Now $1 = \gamma(c)= \gamma(\varphi_1(b))$ $= (\gamma \circ \varphi_1)(b)$ $= (\varphi_2 \circ \beta)(b)$ $= \varphi_2(\beta(b))$, so that $\beta(b) \in \mathsf{ker}\ \varphi_2$. Thus $\beta(b) \in \mathsf{im}\ \psi_2$. Say $a^\prime \in A_2$ with $\beta(b) = \psi_2(a^\prime)$. Since $\alpha$ is surjective, there exists $a \in A_1$ such that $\alpha(a) = a^\prime$. Now $\beta(b) = (\psi_2 \circ \alpha)(a) = (\beta \circ \psi_1)(a)$ $= \beta(\psi_1(a))$. Since $\beta$ is injective, we have $b = \psi_1(a)$. In particular, $b \in \mathsf{im}\ \psi_1$, so that $b \in \mathsf{ker}\ \varphi_1$. Thus $c = \varphi_1(b) = 1$. So $\mathsf{ker}\ \gamma = 1$, and so $\gamma$ is injective.
2. Let $b \in B_2$. Now $\varphi_2(b) \in C_2$. Since $\gamma$ is surjective, there exists $c^\prime \in C_1$ such that $\gamma(c^\prime) = \varphi_2(b)$. Now $\gamma(c^\prime) \in \mathsf{im}\ \varphi_2$, so that $\gamma(c^\prime) \in \mathsf{ker}\ \chi_2$. Now $(\chi_2 \circ \gamma)(c^\prime) = 1$, so that $(\delta \circ \chi_1)(c^\prime) = 1$. Since $\delta$ is injective, $\chi_1(c^\prime) = 1$. Thus $c^\prime \in \mathsf{ker}\ \chi_1$. So $c^\prime \in \mathsf{im}\ \varphi_1$. Say $b^\prime \in B_1$ such that $\varphi_1(b^\prime) = c^\prime$. Now $\varphi_2(b) = \gamma(c^\prime)$ $= (\gamma \circ \varphi_1)(b^\prime)$ $= (\varphi_2 \circ \beta)(b^\prime)$ $= \varphi_1(\beta(b^\prime))$. Thus $\varphi_2(b^{-1}\beta(b^\prime)) = 1$, so that $b^{-1}\beta(b^\prime) \in \mathsf{ker}\ \varphi_2$. Now $b^{-1}\beta(b^\prime) \in \mathsf{im}\ \psi_2$; say $a \in A_2$ such that $b^{-1}\beta(b^\prime) = \psi_2(a)$. Since $\alpha$ is surjective, there exists $a^\prime \in A_1$ such that $\alpha(a^\prime) = a$. Now $b^{-1}\beta(b^\prime) = (\psi_2 \circ \alpha)(a^\prime)$ $= (\beta \circ \psi_1)(a^\prime)$. Thus $b^{-1}\beta(b^\prime) = \beta(\psi_1(a^\prime))$, and so $b = \beta(b^\prime \psi_1((a^\prime)^{-1}))$. Thus $\beta$ is surjective.

### Some properties of a commutative diagram with exact rows

Consider the following commutative diagram of groups.

A commutative diagram of groups

Suppose the rows are exact; that is, $\mathsf{im}\ \psi_k = \mathsf{ker}\ \varphi_k$ for $k \in \{1,2\}$. Prove the following.

1. If $\varphi_1$ and $\alpha$ are surjective and $\beta$ is injective, then $\gamma$ is injective.
2. If $\psi_2$, $\alpha$, and $\gamma$ are injective, then $\beta$ is injective.
3. If $\varphi_1$, $\alpha$, and $\gamma$ are surjective, then $\beta$ is surjective.
4. If $\beta$ is surjective and $\gamma$ and $\psi_2$ are injective, then $\alpha$ is surjective.

1. Let $c \in \mathsf{ker}\ \gamma$. Since $\varphi_1$ is surjective, there exists $b \in B_1$ such that $\varphi(b) = c$. Now $(\gamma \circ \varphi_1)(b) = 1$, so that $(\varphi_2 \circ \beta)(b) = 1$, and so $\varphi_2(\beta(b)) = 1$. Thus $\beta(b) \in \mathsf{ker}\ \varphi_2$, and so $\beta(b) \in \mathsf{im}\ \psi_2$. Say $\beta(b) = \psi_2(a^\prime)$ where $a^\prime \in A_2$. Since $\alpha$ is surjective, there exists $a \in A_1$ such that $\alpha(a) = a^\prime$. Now $\beta(b) = (\psi_2 \circ \alpha)(a)$, so that $\beta(b) = (\beta \circ \psi_1)(a)$, and so $\beta(b) = \beta(\psi_1(a))$. Since $\beta$ is injective, $b = \psi_1(a)$. In particular, $b \in \mathsf{im}\ \psi_1$, so that $b \in \mathsf{ker}\ \varphi_1$. Then $c = \varphi_1(b) = 1$. Thus $\mathsf{ker}\ \gamma = 1$, and so $\gamma$ is injective.
2. Let $b \in \mathsf{ker}\ \beta$. Now $\beta(b) = 1$, so that $(\varphi_2 \circ \beta)(b) = 1$, so that $(\gamma \circ \varphi_1)(b) = 1$. Thus $\gamma(\varphi_1(b)) = 1$. Since $\gamma$ is injective, $\varphi_1(b) = 1$, so that $b \in \mathsf{ker}\ \varphi_1$. So $b \in \mathsf{im}\ \psi_1$. Thus there exists $a \in A_1$ such that $b = \psi_1(a)$. Now $(\psi_2 \circ \alpha)(a) = (\beta \circ \psi_1)(a) = \beta(\psi_1(a))$ $= \beta(b) = 1$, so that $a \in \mathsf{ker}\ \psi_2 \circ \alpha$. Since $\alpha$ and $\psi_2$ are injective, $\psi_2 \circ \alpha$ is injective, so that $a = 1$. Thus $b = \psi_1(1) = 1$, and thus $\mathsf{ker}\ \beta = 1$. Thus $\beta$ is injective.
3. Let $b \in B_2$. Now $\varphi_2(b) \in C_2$. Since $\gamma$ and $\varphi_1$ are surjective, $\gamma \circ \varphi_1$ is surjective, so that there exists $b^\prime \in B_1$ such that $\varphi_2(b) = (\gamma \circ \varphi_1)(b^\prime)$. Note that $\varphi_2(b^{-1}\beta(b^\prime)) = 1$, so that $b^{-1} \beta(b^\prime) \in \mathsf{ker}\ \varphi_2$. Thus $b^{-1}\beta(b^\prime) \in \mathsf{im}\ \psi_2$, and so there exists $a^\prime \in A_2$ such that $b^{-1}\beta(b^\prime) = \psi_2(a^\prime)$. Since $\alpha$ is surjective, there exists $a \in A_1$ such that $\alpha(a) = a^\prime$. Now $b^{-1}\beta(b^\prime) = (\psi_2 \circ \alpha)(a) = (\beta \circ \psi_1)(a) = \beta(\psi_1(a))$. Thus $b = \beta(b^\prime \psi_1(a^{-1}))$, and so $\beta$ is surjective.
4. Let $a \in A_2$. Now $\psi_2(a) \in B_2$. Since $\beta$ is surjective, there exists $b^\prime \in B_1$ such that $\psi_2(a) = \beta(b^\prime)$. Now $1 = (\varphi_2 \circ \psi_2)(a) = \varphi_2(\psi_2(a))$ $= \varphi_2(\beta(b^\prime))$ $= (\varphi_2 \circ \beta)(b^\prime)$ $= (\gamma \circ \varphi_1)(b^\prime)$ $= \gamma(\varphi_1(b^\prime))$. Since $\gamma$ is injective, $\varphi_1(b^\prime) = 1$. So $b^\prime \in \mathsf{ker}\ \varphi_1$, and so $b^\prime \in \mathsf{im}\ \psi_1$. Say $a^\prime \in A_1$ such that $b^\prime = \psi_1(a^\prime)$. Now $\psi_2(a) = \beta(b^\prime)$ $= \beta(\psi_1(a^\prime))$ $= (\beta \circ \psi_1)(a^\prime)$ $= (\psi_2 \circ \alpha)(a^\prime)$ $= \psi_2(\alpha(a^\prime))$. Since $\psi_2$ is injective, $a = \alpha(a^\prime)$. Thus $\alpha$ is surjective.

### A group homomorphism which is not an R-module homomorphism

Let $R$ be a ring with 1. Give an explicit example of a map from one $R$-module to another which is a group homomorphism but which is not an $R$-module homomorphism.

Let $R$ be any noncommutative ring, and fix $a \in R$ such that $a$ is not in the center of $R$. Consider $M = R$ as a left module over itself via left multiplication.

Note that the mapping $\psi : M \rightarrow M$ given by $\psi(x) = a \cdot x$ is a group homomorphism since $\psi(x + y) = r \cdot (x+y) = r \cdot x + r \cdot y$ $= \psi(x) + \psi(y)$. However, if we choose $s \in R$ not commuting with $r$, then $\psi(s) = r \cdot s = rs$ while $s \cdot \psi(1) = s \cdot (r \cdot 1) = sr$. These two are not equal, so that $\psi$ is not an $R$-module homomorphism.

### Exhibit a mapping which is a group homomorphism but not a ring homomorphism

Let $R$ be the ring of all continuous real valued functions on the closed interval $[0,1]$. Prove that the map $\varphi : R \rightarrow \mathbb{R}$ given by $\varphi(f) = \int_0^1 f(x) dx$ is a homomorphism of additive groups but not a ring homomorphism.

We know from calculus that $\varphi(f+g) = \int_0^1 (f+g)(x) dx$ $= \int_0^1 f(x) + g(x) dx$ $= \int_0^1 f(x) dx + \int_0^1 g(x) dx$ $= \varphi(f) + \varphi(g)$. Thus $\varphi$ is an additive group homomorphism.

On the other hand, note that $\varphi(x^2) = 1/3$, while $\varphi(x)^2 = 1/4$. Thus $\varphi$ is not a ring homomorphism.

### In a semidirect product, the centralizer of the normal factor in the nonnormal factor is the kernel of the defining homomorphism

Let $H$ and $K$ be groups, let $\varphi : K \rightarrow \mathsf{Aut}(H)$ be a homomorphism, and let $G = H \rtimes_\varphi K$. Identify $H = H \times 1$ and $K = 1 \times K$ as subgroups of $G$.

Prove that $C_K(H) = \mathsf{ker}\ \varphi$. (Recall that $C_K(H) = C_G(H) \cap K$.)

$(\subseteq)$ Let $(1,k) \in C_K(H)$. By definition, if $(h,1) \in H \times 1$, then $(h,1) \cdot (1,k) = (1,k) \cdot (h,1)$. By the definition of multiplication in a semidirect product, we have $(h \cdot \varphi(1)(1), k) = (1 \cdot \varphi(k)(h), k)$, so that $(h,k) = (\varphi(k)(h),k)$. Comparing first coordinates, $h = \varphi(k)(h)$. Since $h \in H$ is arbitrary, we have $\varphi(k) = \mathsf{id}_H$, so that $k \in \mathsf{ker}\ \varphi$.

$(\supseteq)$ Suppose $(1,k) \in \mathsf{ker}\ \varphi \leq K$. Then $\varphi(k) = \mathsf{id}_H$. Now let $(h,1) \in H \times 1$ be arbitrary. Then $(1,k) \cdot (h,1) = (\varphi(k)(h), k)$ $= (h,k)$ $= (h \cdot \varphi(1)(1), k)$ $= (h,1) \cdot (1,k)$. Thus $(1,k) \in C_K(H)$.

### A criterion for the existence of unique group homomorphisms from an abelian group

Let $A = \langle x_1 \rangle \times \cdots \langle x_t \rangle$ be a finite abelian group with $|x_i| = n_i$ for each $i$.

1. Find a presentation for $A$.
2. Prove that if $G$ is a group containing commuting elements $g_1,\ldots,g_t$ such that $g_i^{n_i} = 1$ for each $i$, then there is a unique group homomorphism $\theta : A \rightarrow G$ such that $\theta(x_i) = g_i$.

1. We claim that $A \cong B = \langle a_1, \ldots, a_t \ |\ a_i^{n_i} = 1, a_ia_j = a_ja_i \rangle$, where $1 \leq i,j \leq t$. Indeed, the “standard basis” elements $e_i$, consisting of $x_i$ in the $i$-th coordinate and 1 in all other coordinates, satisfy these relations.
2. Note that every element of $A$ can be written uniquely as $(x_i^{b_i})_{i=1}^t$. Define $\theta((x_i^{b_i})) = \prod_{i=1}^t g_i^{b_i}$, where the product on the right hand side is as computed inside $G$. It is clear that $\theta(x_i) = g_i$ for each $i$. Moreover, $\theta$ is a homomorphism because the $g_i$ commute pairwise.

Finally, suppose $\sigma$ is a homomorphism $A \rightarrow G$ such that $\sigma(x_i) = g_i$ for each $i$. Then $\sigma((x_i^{b_i})) = \prod \sigma(x_i)^{b_i}$ $= \prod \theta(x_i)^{b_i}$ $= \theta((x_i^{b_i}))$. Thus $\sigma = \theta$ (products computed inside $G$), so that $\theta$ is unique.

### 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.

### Compute the kernel and fibers of a given group homomorphism

Let $G$ be the additive group of real numbers and $H$ the multiplicative group of complex numbers with absolute value 1. Let $\varphi : G \rightarrow H$ be the homomorphism $r \mapsto e^{4 \pi i r}$. Find the kernel of $\varphi$ and the fibers over $-1$, $i$, and $e^{4 \pi i / 3}$.

Lemma: Let $\alpha, \beta$ be set mappings such that $\alpha \circ \beta$ exists, and let $\alpha^\star$ denote preimages. Then $(\alpha \circ \beta)^\star [A] = \beta^\star [\alpha^\star [A]]$. Proof: If $x \in (\alpha \circ \beta)^\star [A]$, then $(\alpha \circ \beta)(x) \in A$, so that $\alpha(\beta(x)) \in A$. Thus $\beta(x) \in \alpha^\star[A]$, so that $x \in \beta^\star [\alpha^\star [A]]$. If $x \in \beta^\star [\alpha^\star [A]]$, then $\beta(x) \in \alpha^\star[A]$, so that $\alpha(\beta(x)) \in A$, hence $x \in (\alpha \circ \beta)^\star [A]$. $\square$

Note that $\varphi = \overline{\varphi} \circ \tau$, where $\overline{\varphi}$ is the mapping considered in the previous exercise and $\tau(x) = 2x$. Using the lemma, we see that $\mathsf{ker}\ \varphi = \{ n/2 \ |\ n \in \mathbb{Z} \}$, $\varphi^\star(-1) = \{ n/2 + 1/4 \ |\ n \in \mathbb{Z} \}$, $\varphi^\star(i) = \{ n/2 + 1/8 \ |\ n \in \mathbb{Z} \}$, and $\varphi^\star(e^{4 \pi / 3}) = \{ n/2 + 1/3 \ |\ n \in \mathbb{Z} \}$.