To find the left cosets and right cosets of the subgroup \( 4\mathbb{Z} \) in \( 2\mathbb{Z} \), we begin by identifying the sets involved:
- \( 2\mathbb{Z} \) is the set of all even integers, i.e., \( 2\mathbb{Z} = { 2k \mid k \in \mathbb{Z} } \).
- \( 4\mathbb{Z} \) is the set of all integers that are multiples of 4, i.e., \( 4\mathbb{Z} = { 4m \mid m \in \mathbb{Z} } \).
Now, to find the cosets:
Left Cosets
A left coset of \( 4\mathbb{Z} \) in \( 2\mathbb{Z} \) can be expressed as \( a + 4\mathbb{Z} \), where \( a \) is an arbitrary element of \( 2\mathbb{Z} \). Since every element in \( 2\mathbb{Z} \) can be written as \( 2k \) for some integer \( k \), we can consider the distinct cosets generated by choosing representatives from \( 2\mathbb{Z} \).
Let's compute the left cosets for representative values of \( a \):
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For \( a = 0 \): \[ 0 + 4\mathbb{Z} = 4\mathbb{Z} = { 0, 4, -4, 8, -8, \ldots } \]
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For \( a = 2 \): \[ 2 + 4\mathbb{Z} = { 2 + 4m \mid m \in \mathbb{Z} } = { 2, 6, -2, 10, -6, \ldots } \]
No other choices of \( a \) in \( 2\mathbb{Z} \) (like \( 4 \) or \( -2 \)) will give new cosets since they will be equivalent to either \( 0 + 4\mathbb{Z} \) or \( 2 + 4\mathbb{Z} \).
Thus, the left cosets are:
- \( 4\mathbb{Z} \)
- \( 2 + 4\mathbb{Z} \)
Right Cosets
Similarly, the right cosets of \( 4\mathbb{Z} \) in \( 2\mathbb{Z} \) are given by \( 4\mathbb{Z} + a \).
Calculating for the same representatives:
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For \( a = 0 \): \[ 4\mathbb{Z} + 0 = 4\mathbb{Z} = { 0, 4, -4, 8, -8, \ldots } \]
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For \( a = 2 \): \[ 4\mathbb{Z} + 2 = { 4m + 2 \mid m \in \mathbb{Z} } = { 2, 6, -2, 10, -6, \ldots } \]
Again, there's no new coset from any other even integer besides \( 0 \) and \( 2 \).
The right cosets are thus also:
- \( 4\mathbb{Z} \)
- \( 2 + 4\mathbb{Z} \)
Summary
The left and right cosets of the subgroup \( 4\mathbb{Z} \) in \( 2\mathbb{Z} \) are:
- \( 4\mathbb{Z} \)
- \( 2 + 4\mathbb{Z} \)
Both the left and right cosets are the same in this case, reflecting the abelian property of the group presented here.