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.
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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.
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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.
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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.
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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.
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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 \(2 \times 2 = 4\).
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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 \(\mathbb{Z}_6\).