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.

Answer 1

To determine if the direct product \((\mathbb{Z}_4 / M) \times (\mathbb{Z}_6 / N)\) forms a group, we first need to analyze the subgroups \(M\) and \(N\) of the respective groups \(\mathbb{Z}_4\) and \(\mathbb{Z}_6\).

1. **Subgroup \(M = \langle 2 \rangle \subset \mathbb{Z}_4\):**
- The group \(\mathbb{Z}_4\) consists of the elements \(\{0, 1, 2, 3\}\).
- The subgroup generated by \(2\) is:
\[
M = \langle 2 \rangle = \{ 0, 2 \}
\]
- Thus, the quotient group \(\mathbb{Z}_4 / M\) consists of the cosets:
\[
\mathbb{Z}_4 / M = \{ M, 1 + M, 3 + M \} = \{ \{0, 2\}, \{1, 3\} \}
\]
- Therefore, \(\mathbb{Z}_4 / M\) has order \(2\) and is isomorphic to \(\mathbb{Z}_2\).

2. **Subgroup \(N = \langle 2 \rangle \subset \mathbb{Z}_6\):**
- The group \(\mathbb{Z}_6\) consists of the elements \(\{0, 1, 2, 3, 4, 5\}\).
- The subgroup generated by \(2\) is:
\[
N = \langle 2 \rangle = \{ 0, 2, 4 \}
\]
- Thus, the quotient group \(\mathbb{Z}_6 / N\) consists of the cosets:
\[
\mathbb{Z}_6 / N = \{ N, 1 + N, 3 + N, 5 + N \} = \{ \{0, 2, 4\}, \{1, 3, 5\} \}
\]
- Therefore, \(\mathbb{Z}_6 / N\) has order \(2\) and is also isomorphic to \(\mathbb{Z}_2\).

3. **Direct Product \((\mathbb{Z}_4 / M) \times (\mathbb{Z}_6 / N)\):**
- Since both \(\mathbb{Z}_4 / M\) and \(\mathbb{Z}_6 / N\) are isomorphic to \(\mathbb{Z}_2\), their direct product is:
\[
\mathbb{Z}_2 \times \mathbb{Z}_2
\]
- This group consists of the elements \((0,0), (0,1), (1,0), (1,1)\).

Thus, we conclude that the direct product \((\mathbb{Z}_4 / M) \times (\mathbb{Z}_6 / N)\) forms a group, specifically isomorphic to \(\mathbb{Z}_2 \times \mathbb{Z}_2\).