Let’s answer each question step by step, as you requested. We'll use the **first isomorphism theorem** to construct a homomorphism with a specific kernel, and then apply the **fourth isomorphism theorem** and draw the lattice diagram. --- ## **1. Q₈ with kernel ⟨-1⟩** ### **Step 1: Find a homomorphism φ: Q₈ → G with kernel ⟨-1⟩** Recall: - The quaternion group \( Q_8 = \{1, -1, i, -i, j, -j, k, -k\} \) - ⟨-1⟩ = {1, -1} is a normal subgroup of Q₈. We want a group homomorphism φ from Q₈ onto some group G such that **ker(φ) = ⟨-1⟩**. #### **Solution:** - The quotient group \( Q_8 / \langle -1 \rangle \) has order 4 and is isomorphic to the Klein four-group \( V_4 \cong C_2 \times C_2 \). - By the **First Isomorphism Theorem**: \( Q_8 / \ker(\phi) \cong \text{im}(\phi) \). - So, we want \( \operatorname{im}(\phi) \cong V_4 \). Define φ as the canonical projection: \[ \phi : Q_8 \to Q_8 / \langle -1 \rangle \] with: - **Kernel:** \( \ker(\phi) = \langle -1 \rangle \) - **Image:** \( Q_8 / \langle -1 \rangle \cong V_4 \) - **Explicit map:** \( \phi(g) = g\langle -1 \rangle \) (the coset of g) **End of Step 1.** --- ### **Step 2: Use the fourth isomorphism theorem and draw the lattice** - The **Fourth Isomorphism Theorem** says there is a bijection between the subgroups of Q₈ containing ⟨-1⟩ and the subgroups of \( Q_8 / \langle -1 \rangle \). - The subgroup lattice above ⟨-1⟩ in Q₈ mirrors the subgroup lattice of \( Q_8 / \langle -1 \rangle \) (which is \( V_4 \)). - \( V_4 \) has three subgroups of order 2, all normal, corresponding to the three subgroups of Q₈ of order 4 containing ⟨-1⟩. #### **Lattice Diagram:** ``` Q₈ / | \ ⟨i⟩ ⟨j⟩ ⟨k⟩ \ | / ⟨-1⟩ | {1} ``` - The top is Q₈, next are ⟨i⟩, ⟨j⟩, ⟨k⟩ (all contain -1), then ⟨-1⟩, then {1}. - In the quotient, this corresponds to V₄ with its three order-2 subgroups. **End of Step 2.** --- ### **Final Answer for Q₈:** - **Homomorphism:** φ: Q₈ → Q₈/⟨-1⟩, φ(g) = g⟨-1⟩ - **Kernel:** ⟨-1⟩ = {1, -1} - **Image:** V₄ (Klein 4-group) - **Lattice:** As above, matching subgroups containing ⟨-1⟩ with those of V₄. --- ## **2. Gaussian Integers G = {1, -1, i, -i} under multiplication, kernel {i}** ### **Step 1: Find a homomorphism φ: G → H with kernel {i}** - G is the group of 4th roots of unity, isomorphic to C₄ (cyclic of order 4). - ⟨i⟩ = {1, i, -1, -i} = G itself, so if the kernel is {i}, it would be {1, i, -i, -1} ∩ {i} = {i}? - But {i} is not a subgroup of G under multiplication, since i·i = -1 ≠ i. - The only subgroups of G are: - {1} - {1, -1} - G itself So, **having kernel {i} is not possible** for a group homomorphism from G, since {i} is not a subgroup. But if the question means that the kernel is ⟨i⟩ (the cyclic subgroup generated by i), then: - ⟨i⟩ = {1, i, -1, -i} = G - In that case, the only homomorphism with kernel G is the trivial homomorphism. #### **Assuming the question meant kernel {1, -1}:** - {1, -1} is a normal subgroup of G of order 2. - G / {1, -1} ≅ C₂ Define φ: G → G / {1, -1} by φ(g) = g {1, -1}. - **Kernel:** {1, -1} - **Image:** C₂ **End of Step 1.** --- ### **Step 2: Fourth isomorphism theorem and lattice diagram** - Subgroups of G containing {1, -1} are {1, -1} and G itself. - G / {1, -1} is C₂, which has two subgroups: {1} and C₂. #### **Lattice Diagram:** ``` G | {1, -1} | {1} ``` - In the quotient, C₂ has {1} < C₂. **End of Step 2.** --- ### **Final Answer for Gaussian Integers:** - **Homomorphism:** φ: G → C₂, φ(g) = g {1, -1} - **Kernel:** {1, -1} - **Image:** C₂ - **Lattice:** As above; subgroups containing {1, -1} correspond to subgroups of C₂. --- ## **Summary Table** | Part | Homomorphism | Kernel | Image | Lattice Diagram Description | |----------|----------------------------------------------|---------------|----------|----------------------------------------------| | Q₈ | φ(g) = g⟨-1⟩ | ⟨-1⟩ | V₄ | Three order-4 subgroups above ⟨-1⟩ | | Gaussian | φ(g) = g {1, -1} | {1, -1} | C₂ | Only {1, -1} and G above {1} | --- **If you need the explicit subgroup correspondences for the fourth isomorphism theorem, or explicit diagram drawings, let me know!**