1. **Problem Statement:** We have a map $f : (G, \Delta) \to (\mathbb{R} \setminus \{0\}, \times)$ defined by $f(x) = x - \alpha$. We want to show that $f$ is a group isomorphism and find its inverse. 2. **Showing $f$ is a group isomorphism:** - A group isomorphism is a bijective homomorphism. - We must show $f(x \Delta y) = f(x) \times f(y)$ for all $x,y \in G$. 3. **Check homomorphism property:** Assuming $\Delta$ is the group operation on $G$, and $\times$ is multiplication on $\mathbb{R} \setminus \{0\}$, we want: $$f(x \Delta y) = f(x) \times f(y)$$ Given $f(x) = x - \alpha$, then $$f(x \Delta y) = (x \Delta y) - \alpha$$ and $$f(x) \times f(y) = (x - \alpha)(y - \alpha)$$ For these to be equal for all $x,y$, the operation $\Delta$ must satisfy: $$(x \Delta y) - \alpha = (x - \alpha)(y - \alpha)$$ Rearranged: $$x \Delta y = (x - \alpha)(y - \alpha) + \alpha$$ This defines the group operation $\Delta$ on $G$. 4. **Bijectivity:** - Since $f(x) = x - \alpha$ is a translation, it is bijective. - The inverse map is $f^{-1}(z) = z + \alpha$ for $z \in \mathbb{R} \setminus \{0\}$. 5. **Answer for part 2 and 3:** - $f$ is a group isomorphism between $(G, \Delta)$ and $(\mathbb{R} \setminus \{0\}, \times)$. - The inverse is $f^{-1}(z) = z + \alpha$. --- **Exercise 2: Subgroup tests** Recall subgroup criteria: nonempty, closed under group operation, and closed under inverses. 1. $G = (\mathbb{Z}, +)$, $H = \{\text{even integers}\}$ - Even integers are closed under addition and inverses. - Identity $0$ is even. - So $H$ is a subgroup. 2. $G = (\mathbb{Z}, +)$, $H = \{\text{odd integers}\}$ - Sum of two odd integers is even, not odd. - Not closed under addition. - So $H$ is not a subgroup. 3. $G = (\mathbb{R}, +)$, $H = [-1, \infty)$ - $H$ contains $-1$ but not $0$ (identity). - Also, not closed under inverses (e.g., $-2 \notin H$). - So $H$ is not a subgroup. 4. $G = (\mathbb{R}^*, \times)$, $H = \mathbb{Q}^*$ - $\mathbb{Q}^*$ is the nonzero rationals. - Closed under multiplication and inverses. - Identity $1$ is rational. - So $H$ is a subgroup. --- **Exercise 3: Ring structure on $\mathbb{Z} \times A$** Given $(A, \star, \Delta)$ is a ring with identity. Define: $$(n,x) \oplus (m,y) = (n+m, x \star y)$$ $$(n,x) \otimes (m,y) = (n \cdot m, x \Delta y)$$ 1. $(\mathbb{Z} \times A, \oplus)$ is a commutative group by assumption. 2. To show $(\mathbb{Z} \times A, \oplus, \otimes)$ is a ring with identity: - Check distributivity: $$(a,b) \otimes ((c,d) \oplus (e,f)) = (a,b) \otimes (c+e, d \star f) = (a(c+e), b \Delta (d \star f))$$ - Using ring distributivity in $A$: $$b \Delta (d \star f) = (b \Delta d) \star (b \Delta f)$$ - Similarly, $$(a,b) \otimes (c,d) \oplus (a,b) \otimes (e,f) = (ac, b \Delta d) \oplus (ae, b \Delta f) = (ac+ae, (b \Delta d) \star (b \Delta f))$$ - Since $a(c+e) = ac + ae$ in $\mathbb{Z}$ and $b \Delta (d \star f) = (b \Delta d) \star (b \Delta f)$ in $A$, distributivity holds. 3. Identity element: - Identity in $\mathbb{Z}$ is $1$. - Identity in $A$ is $e_A$. - So identity in $\mathbb{Z} \times A$ is $(1, e_A)$. 4. Commutativity: - $\oplus$ is commutative by assumption. - $\otimes$ is commutative if and only if both $\cdot$ in $\mathbb{Z}$ and $\Delta$ in $A$ are commutative. - Since $\cdot$ in $\mathbb{Z}$ is commutative, commutativity depends on $\Delta$ in $A$. **Final answers:** - $(\mathbb{Z} \times A, \oplus, \otimes)$ is a ring with identity $(1, e_A)$. - It is commutative if and only if $A$ is commutative under $\Delta$.