Let G denote the multiplicative group z22 and N = 21 denote the subgroup of G. Show that the quotient group G/N is isomorphic to Z5

To show that the quotient group \( G/N \) is isomorphic to \( \mathbb{Z}_5 \), we need to go through a few steps:

### Step 1: Understand the Group \( G = \mathbb{Z}_{22}^* \)

The group \( G \) is the multiplicative group of integers modulo 22, which we denote as \( \mathbb{Z}_{22}^* \). The elements of \( G \) are the integers less than 22 that are coprime to 22, meaning they have no common divisors with 22 other than 1.

The elements of \( G \) are:
\[
G = \{1, 3, 5, 7, 9, 13, 15, 17, 19, 21\}
\]
This group has 10 elements.

### Step 2: Identify the Subgroup \( N = \{21\} \)

Here, the subgroup \( N \) is generated by the element 21 in \( G \). Since \( 21 \times 21 \equiv 1 \ (\text{mod} \ 22) \), the subgroup \( N \) is simply:
\[
N = \{1, 21\}
\]
This subgroup contains two elements and is cyclic of order 2.

### Step 3: Determine the Quotient Group \( G/N \)

The quotient group \( G/N \) consists of the cosets of \( N \) in \( G \). A coset of \( N \) in \( G \) is of the form \( gN = \{gn \mid n \in N\} \) for some \( g \in G \). Given that \( N = \{1, 21\} \), the cosets are:

1. \( 1N = \{1, 21\} \)
2. \( 3N = \{3, 3 \times 21 \equiv 17 \ (\text{mod} \ 22)\} \)
3. \( 5N = \{5, 5 \times 21 \equiv 17 \ (\text{mod} \ 22)\} \)
4. \( 7N = \{7, 7 \times 21 \equiv 15 \ (\text{mod} \ 22)\} \)
5. \( 9N = \{9, 9 \times 21 \equiv 13 \ (\text{mod} \ 22)\} \)

Notice that these cosets can be written as:
\[
G/N = \{1N, 3N, 5N, 7N, 9N\}
\]
There are 5 distinct cosets, and thus \( G/N \) has 5 elements.

### Step 4: Isomorphism with \( \mathbb{Z}_5 \)

The group \( \mathbb{Z}_5 \) is the additive group of integers modulo 5, which has 5 elements: \( \{0, 1, 2, 3, 4\} \). To show that \( G/N \) is isomorphic to \( \mathbb{Z}_5 \), we need to establish a bijective homomorphism.

Define a mapping \( \phi: G/N \to \mathbb{Z}_5 \) by:
\[
\phi(gN) = k \ (\text{mod} \ 5) \ \text{where} \ g = 3^k \ (\text{mod} \ 22)
\]

Let's verify that this map is a homomorphism:
- \( \phi(1N) = 0 \)
- \( \phi(3N) = 1 \)
- \( \phi(5N) = 2 \)
- \( \phi(7N) = 3 \)
- \( \phi(9N) = 4 \)

### Step 5: Verify that \( \phi \) is an Isomorphism

1. **Homomorphism:** We check that the operation is preserved. For example:
   \[
   \phi(3N \cdot 3N) = \phi(9N) = 4 \equiv 1 + 1 \ (\text{mod} \ 5)
   \]

2. **Bijective:** \( \phi \) is injective (one-to-one) because each coset in \( G/N \) maps to a unique element in \( \mathbb{Z}_5 \). It is also surjective (onto) because every element in \( \mathbb{Z}_5 \) has a preimage in \( G/N \).

Since \( \phi \) is a bijective homomorphism, it is an isomorphism.

### Conclusion

The quotient group \( G/N \) is isomorphic to \( \mathbb{Z}_5 \).

Would you like to explore details of this concept or any other related topic?

### Related Questions
1. What is the significance of quotient groups in group theory?
2. How do you construct the cosets of a subgroup in a given group?
3. What are some common examples of quotient groups in algebra?
4. How can we determine the order of an element in a quotient group?
5. What is the role of isomorphisms in the classification of groups?

### Tip
Always verify both injectivity and surjectivity when proving that a mapping is an isomorphism.

Learn how to demonstrate that the quotient group G/N, where G is the multiplicative group Z22 and N = {21}, is isomorphic to the cyclic group Z5. Understand the steps involved in constructing G/N, identifying cosets, and establishing a bijective homomorphism to Z5. Suitable for advanced undergraduate students and enthusiasts of group theory.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation