Question

Let 𝑁 = < 2 > be a subgroup of (ℤ6, +) and 𝑀 = < 2 > be a subgroup of (ℤ4, +) and direct product of
(ℤ4 /𝑀) × (ℤ6 /𝑁) forms a group.
a) Show whether the direct product of (ℤ4 /𝑀) × (ℤ6 /𝑁) is isomorphic to ℤ6

Answer 1

To analyze whether the direct product \((\mathbb{Z}_4 / M) \times (\mathbb{Z}_6 / N)\) is isomorphic to \(\mathbb{Z}_6\), we start by determining the structures of the relevant groups and their respective quotients.

1. **Find \(M = \langle 2 \rangle\) in \(\mathbb{Z}_4\)**:
- The group \(\mathbb{Z}_4\) consists of the elements \(\{0, 1, 2, 3\}\).
- The subgroup generated by 2 is:
\[
M = \{0, 2\} \text{ (i.e., } \langle 2 \rangle = \{0, 2\}\text{)}
\]
- The order of \(M\) is 2.

2. **Compute the quotient \(\mathbb{Z}_4 / M\)**:
- The cosets of \(M\) in \(\mathbb{Z}_4\) are:
\[
M = \{0, 2\}
\]
\[
1 + M = \{1, 3\}
\]
- Therefore,
\[
\mathbb{Z}_4 / M = \{M, 1 + M\} = \{\{0, 2\}, \{1, 3\}\}
\]
- The quotient \(\mathbb{Z}_4 / M\) has 2 elements.

3. **Find \(N = \langle 2 \rangle\) in \(\mathbb{Z}_6\)**:
- The group \(\mathbb{Z}_6\) consists of the elements \(\{0, 1, 2, 3, 4, 5\}\).
- The subgroup generated by 2 is:
\[
N = \{0, 2, 4\}
\text{ (i.e., } \langle 2 \rangle = \{0, 2, 4\}\text{)}
\]
- The order of \(N\) is 3.

4. **Compute the quotient \(\mathbb{Z}_6 / N\)**:
- The cosets of \(N\) in \(\mathbb{Z}_6\) are:
\[
N = \{0, 2, 4\}
\]
\[
1 + N = \{1, 3, 5\}
\]
- Therefore,
\[
\mathbb{Z}_6 / N = \{N, 1 + N\} = \{\{0, 2, 4\}, \{1, 3, 5\}\}
\]
- The quotient \(\mathbb{Z}_6 / N\) also has 2 elements.

5. **Analyze the direct product \((\mathbb{Z}_4 / M) \times (\mathbb{Z}_6 / N)\)**:
- We have:
\[
\mathbb{Z}_4 / M \cong \mathbb{Z}_2 \quad (\text{2 elements})
\]
\[
\mathbb{Z}_6 / N \cong \mathbb{Z}_2 \quad (\text{2 elements})
\]
- The direct product \((\mathbb{Z}_2) \times (\mathbb{Z}_2)\) has order \(2 \times 2 = 4\).

6. **Determine whether \((\mathbb{Z}_2 \times \mathbb{Z}_2)\) is isomorphic to \(\mathbb{Z}_6\)**:
- The group \(\mathbb{Z}_6\) has 6 elements:
\[
\mathbb{Z}_6 = \{0, 1, 2, 3, 4, 5\}
\]
- Since \(\mathbb{Z}_6\) is cyclic and has an element of order 6, and \((\mathbb{Z}_2 \times \mathbb{Z}_2)\) is non-cyclic (every element has order 2), we conclude that they cannot be isomorphic.

Thus, \((\mathbb{Z}_4 / M) \times (\mathbb{Z}_6 / N)\) is not isomorphic to \(\mathbb{Z}_6\).