1. **State the problem:** Determine which of the given subgroup relations are correct. 2. **Recall the definition:** For groups $H$ and $G$, $H \leq G$ means $H$ is a subgroup of $G$. 3. **Analyze each relation:** - a. $\mathbb{Z}_2 \leq \mathbb{Z}_4$? $\mathbb{Z}_2$ is the cyclic group of order 2, $\mathbb{Z}_4$ is cyclic of order 4. Since $\mathbb{Z}_4$ has a unique subgroup of order 2 isomorphic to $\mathbb{Z}_2$, this is true. - b. $\mathbb{Z}_4 \leq \mathbb{Z}$? $\mathbb{Z}$ is infinite cyclic, $\mathbb{Z}_4$ is finite cyclic. A finite cyclic group cannot be a subgroup of an infinite cyclic group because subgroup orders divide the group order, and infinite order cannot contain finite order elements except the trivial subgroup. So this is false. - c. $2\mathbb{Z} \leq 4\mathbb{Z}$? $2\mathbb{Z} = \{2k : k \in \mathbb{Z}\}$, $4\mathbb{Z} = \{4k : k \in \mathbb{Z}\}$. Every element of $4\mathbb{Z}$ is in $2\mathbb{Z}$ but not vice versa. So $2\mathbb{Z}$ is not a subset of $4\mathbb{Z}$, so false. - d. $4\mathbb{Z} \leq 2\mathbb{Z}$? Every element of $4\mathbb{Z}$ is in $2\mathbb{Z}$ since $4k = 2(2k)$. So $4\mathbb{Z}$ is a subgroup of $2\mathbb{Z}$, true. **Final answer:** a and d are correct relations.