give answer in 2 step with explanation at the end of each step and final answer at the end:

1
-
Using the first isomorphism theorem, find an homomorphism, kernel of phi, image of a map from Q
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Question

give answer in 2 step with explanation at the end of each step and final answer at the end:

1
-
 ﻿Using the first isomorphism theorem, find an homomorphism, kernel of phi, image of a map from Q
8
 ﻿such the kernel is 
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-
1
>
?
 ﻿Please select proper image. Use also the fourth isomorphism theorem to explain and hence draw the corresponding lattice diagram. 
2
-
 ﻿Using the first isomorphism theorem, find an homomorphism, kernel of phi, image of a map from the set of Gaussian integers G
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 ﻿under multiplication such the kernel is 
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i
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 ﻿Please select proper image. Use also the fourth isomorphism theorem to explain and hence draw the corresponding lattice diagram

VIPSolutions · Accepted Answer

Let's tackle each part step by step as you requested.

---

### **1. Q₈ and Kernel ⟨-1⟩**

#### **Step 1: Homomorphism, Kernel, and Image**

- **Group:** $ Q_8 = \{\pm1, \pm i, \pm j, \pm k\} $ (the quaternion group)
- **Subgroup:** $ \langle -1 \rangle = \{1, -1\} $

We want a homomorphism $ \phi: Q_8 \to G $ such that $ \ker(\phi) = \langle -1 \rangle $.

**Natural Choice:**
- The quotient group $ Q_8 / \langle -1 \rangle $ has order 4.
- $ Q_8 / \langle -1 \rangle \cong V_4 $ (the Klein 4-group), since quotienting by the center in $ Q_8 $ yields $ V_4 $.

**Homomorphism:**
- Define $ \phi: Q_8 \to V_4 $ by mapping each element to its coset in $ Q_8 / \langle -1 \rangle $.

**Kernel:** $ \langle -1 \rangle $, by construction.

**Image:** All of $ V_4 $, since this is the quotient.

**Explanation (Step 1):**  
The quotient map $ \phi: Q_8 \to Q_8 / \langle -1 \rangle $ is a surjective homomorphism with kernel $ \langle -1 \rangle $, and the image is $ V_4 $.

---

#### **Step 2: Fourth Isomorphism Theorem & Lattice**

- The **Fourth Isomorphism Theorem** says there is a lattice isomorphism between subgroups of $ Q_8 $ containing $ \langle -1 \rangle $ and subgroups of $ V_4 $.

**Subgroups of $ Q_8 $ containing $ \langle -1 \rangle $:**
- $ \langle -1 \rangle $
- $ \langle i, -1 \rangle = \{\pm 1, \pm i\} $ (same for $ j $, $ k $)
- $ Q_8 $

**Subgroups of $ V_4 $:**
- $ \{e\} $
- Three subgroups of order 2: $ \langle a \rangle, \langle b \rangle, \langle ab \rangle $
- $ V_4 $

**Lattice Diagram:**
```
     Q8                  V4
      |                   |
           
      |                   |
 (kernel)           {e}
```

**Explanation (Step 2):**  
The fourth isomorphism theorem gives a matching between subgroups of $ Q_8 $ above $ \langle -1 \rangle $ and subgroups of $ V_4 $, showing how the subgroup structure "descends" via the quotient.

---

**Final Answer for 1:**

- **Homomorphism:** The natural quotient map $ \phi: Q_8 \to Q_8 / \langle -1 \rangle \cong V_4 $.
- **Kernel:** $ \langle -1 \rangle $
- **Image:** $ V_4 $
- **Lattice:** Subgroups above $ \langle -1 \rangle $ in $ Q_8 $ correspond to subgroups of $ V_4 $.

---

### **2. Gaussian Integers $ G = \{1, -1, i, -i\} $ with Kernel $ \langle i \rangle $**

#### **Step 1: Homomorphism, Kernel, and Image**

- $ G = \{1, -1, i, -i\} $, forms the group $ C_4 $ (cyclic of order 4) under multiplication.
- $ \langle i \rangle = \{1, i, -1, -i\} = G $. However, if you mean the subgroup generated by $ i $, that's the whole group.

But, perhaps you mean the kernel is $ \{1, -1\} $ (since $ i $ would generate the whole group).

Suppose **kernel is $ \langle i \rangle = G $:**
- Then the only homomorphism is the trivial map, which is not interesting.

Suppose **kernel is $ \{1, -1\} $:**
- Then $ G / \langle -1 \rangle \cong C_2 $.
- Define $ \phi: G \to C_2 $ by $ \phi(z) = z^2 $. Check:
  - $ (1)^2 = 1 $, $ (-1)^2 = 1 $, $ i^2 = -1 $, $ (-i)^2 = -1 $.
- So $ \phi(1) = 1 $, $ \phi(-1) = 1 $, $ \phi(i) = -1 $, $ \phi(-i) = -1 $.
- Kernel is $ \{1, -1\} $, image is $ \{1, -1\} $ (i.e., $ C_2 $).

**Explanation (Step 1):**  
The map $ \phi: G \to C_2 $ given by $ \phi(z) = z^2 $ is a surjective homomorphism, kernel $ \{1, -1\} $, image $ C_2 $.

---

#### **Step 2: Fourth Isomorphism Theorem & Lattice**

- Subgroups of $ G $ containing $ \{1, -1\} $: $ \{1, -1\} $, $ G $.
- Subgroups of $ C_2 $: $ \{1\} $, $ C_2 $.

**Lattice Diagram:**
```
    G                C2
    |                 |
 {1,-1}             {1}
```

**Explanation (Step 2):**  
The fourth isomorphism theorem shows subgroups of $ G $ containing $ \{1, -1\} $ correspond to subgroups of $ C_2 $.

---

**Final Answer for 2:**

- **Homomorphism:** $ \phi(z) = z^2 $, where $ \phi: G \to C_2 $.
- **Kernel:** $ \{1, -1\} $
- **Image:** $ C_2 $
- **Lattice:** Subgroups of $ G $ containing $ \{1, -1\} $ correspond to subgroups of $ C_2 $.

---

If you specifically want the kernel to be $ \langle i \rangle $ (which is all of $ G $), only the trivial map works.  
But usually "kernel is $ i $" means subgroup generated by $ i $, which is the whole group in this context.  
So, the answer above assumes you meant kernel $ \{1, -1\} $. If you want the trivial map (kernel $ G $), let me know!

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