1. **Problem (a): Determine the order of the group $G=\mathbb{Z}_6$ and the order of a subgroup in $G$.** The group $\mathbb{Z}_6$ consists of integers modulo 6 under addition: $\{0,1,2,3,4,5\}$. - The order of a group is the number of elements in the group. - Here, $|G| = 6$ because there are 6 elements. Next, consider a subgroup $H$ of $G$. For example, the subgroup generated by 2 is $H = \{0,2,4\}$. - The order of this subgroup $H$ is the number of elements in $H$, which is 3. 2. **Problem (b): Validate that the set of non-zero real numbers $\mathbb{R}^*$ with multiplication is a commutative group.** To verify this, we check the group axioms: 1. **Closure:** For any $a,b \in \mathbb{R}^*$, $a \times b \in \mathbb{R}^*$ since the product of two non-zero real numbers is non-zero. 2. **Associativity:** Multiplication of real numbers is associative: $a \times (b \times c) = (a \times b) \times c$. 3. **Identity element:** The number 1 is the identity since $a \times 1 = a$ for all $a \in \mathbb{R}^*$. 4. **Inverse element:** For each $a \in \mathbb{R}^*$, the inverse is $a^{-1} = \frac{1}{a}$, which is also in $\mathbb{R}^*$. 5. **Commutativity:** Multiplication of real numbers is commutative: $a \times b = b \times a$. Thus, $\mathbb{R}^*$ with multiplication is a commutative group. **Final answers:** - Order of $G=\mathbb{Z}_6$ is $6$. - Order of subgroup $H=\{0,2,4\}$ is $3$. - $\mathbb{R}^*$ with multiplication is a commutative group.