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 determine whether \((\mathbb{Z}_4 / M) \times (\mathbb{Z}_6 / N)\) is isomorphic to \(\mathbb{Z}_6\), we first need to identify the subgroups \(M\) and \(N\) and find the quotient groups.

### Step 1: Find \(M\) and \(N\)

1. **Subgroup \(N = \langle 2 \rangle\) in \((\mathbb{Z}_6, +)\)**:
- The elements of \(\mathbb{Z}_6 = \{0, 1, 2, 3, 4, 5\}\).
- The subgroup generated by \(2\) consists of:
\[
\{0, 2, 4\} = N
\]
- The order of \(N\) is \(3\) (elements are \(0, 2, 4\)).

2. **Subgroup \(M = \langle 2 \rangle\) in \((\mathbb{Z}_4, +)\)**:
- The elements of \(\mathbb{Z}_4 = \{0, 1, 2, 3\}\).
- The subgroup generated by \(2\) consists of:
\[
\{0, 2\} = M
\]
- The order of \(M\) is \(2\) (elements are \(0, 2\)).

### Step 2: Compute the quotient groups

1. **Quotient \(\mathbb{Z}_4 / M\)**:
- \(\mathbb{Z}_4 / M\) has \( \frac{4}{2} = 2 \) elements.
- The elements of \(\mathbb{Z}_4 / M\) are represented by the cosets:
- \(M = \{0, 2\}\)
- \(1 + M = \{1, 3\}\)
- Therefore, \(\mathbb{Z}_4 / M = \{\{0, 2\}, \{1, 3\}\} \cong \mathbb{Z}_2\).

2. **Quotient \(\mathbb{Z}_6 / N\)**:
- \(\mathbb{Z}_6 / N\) has \( \frac{6}{3} = 2 \) elements.
- The elements of \(\mathbb{Z}_6 / N\) are represented by the cosets:
- \(N = \{0, 2, 4\}\)
- \(1 + N = \{1, 3, 5\}\)
- Therefore, \(\mathbb{Z}_6 / N = \{\{0, 2, 4\}, \{1, 3, 5\}\} \cong \mathbb{Z}_2\).

### Step 3: Form the direct product

Now we compute the direct product:
\[
\mathbb{Z}_4 / M \times \mathbb{Z}_6 / N \cong \mathbb{Z}_2 \times \mathbb{Z}_2
\]
The group \(\mathbb{Z}_2 \times \mathbb{Z}_2\) has \(4\) elements and is often represented as a Klein four-group:
\[
\mathbb{Z}_2 \times \mathbb{Z}_2 = \{(0,0), (0,1), (1,0), (1,1)\}
\]

### Conclusion: Compare with \(\mathbb{Z}_6\)

The group \(\mathbb{Z}_6\) is a cyclic group with \(6\) elements, specifically:
\[
\mathbb{Z}_6 = \{0, 1, 2, 3, 4, 5\}
\]
\(\mathbb{Z}_2 \times \mathbb{Z}_2\) is a group of order \(4\) (and not cyclic, as it has no element of order \(4\)), while \(\mathbb{Z}_6\) has elements of order \(1, 2, 3\).

Thus, since \(\mathbb{Z}_4 / M \times \mathbb{Z}_6 / N\) is \(\mathbb{Z}_2 \times \mathbb{Z}_2\) (with order \(4\)) and \(\mathbb{Z}_6\) has order \(6\), we conclude that:
\[
(\mathbb{Z}_4 / M) \times (\mathbb{Z}_6 / N) \text{ is NOT isomorphic to } \mathbb{Z}_6.
\]