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The set of four rotations of a square of its plane about a normal passing through the centre of the plane form a
- a)group but not abelian
- b)abelian group but not cyclic
- c)cyclic group
- d)symmetric group
Correct answer is option 'C'. Can you explain this answer?
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The set of four rotations of a square of its plane about a normal pass...
Explanation:
To understand why the set of four rotations of a square about a normal passing through the center of the plane forms a cyclic group, we need to understand what a cyclic group is and how the rotations of a square can be represented as a cyclic group.
Cyclic Group:
A cyclic group is a group that can be generated by a single element. It is a group where all the elements can be obtained by repeatedly applying the group operation to a single element. In other words, there exists an element in the group, called the generator, such that every other element in the group can be expressed as a power of the generator.
Rotations of a Square:
A square has four rotational symmetries: no rotation, a 90-degree rotation, a 180-degree rotation, and a 270-degree rotation. These rotations can be represented using matrix multiplication. Let's denote the no rotation as R0, the 90-degree rotation as R1, the 180-degree rotation as R2, and the 270-degree rotation as R3.
The matrix representation of these rotations can be written as follows:
R0 = [1 0]
[0 1]
R1 = [0 -1]
[1 0]
R2 = [-1 0]
[0 -1]
R3 = [0 1]
[-1 0]
Cyclic Group of Rotations:
If we consider the composition of rotations, we can see that rotating the square by 90 degrees four times brings it back to its original position. In other words, R1 * R1 * R1 * R1 = R4 = R0.
Similarly, rotating the square by 90 degrees three times gives us R3, rotating it by 90 degrees twice gives us R2, and rotating it by 90 degrees once gives us R1.
Therefore, if we consider the set {R0, R1, R2, R3}, we can see that it forms a cyclic group generated by R1. Every other rotation in the group can be expressed as a power of R1.
Hence, the set of four rotations of a square about a normal passing through the center of the plane forms a cyclic group.
To understand why the set of four rotations of a square about a normal passing through the center of the plane forms a cyclic group, we need to understand what a cyclic group is and how the rotations of a square can be represented as a cyclic group.
Cyclic Group:
A cyclic group is a group that can be generated by a single element. It is a group where all the elements can be obtained by repeatedly applying the group operation to a single element. In other words, there exists an element in the group, called the generator, such that every other element in the group can be expressed as a power of the generator.
Rotations of a Square:
A square has four rotational symmetries: no rotation, a 90-degree rotation, a 180-degree rotation, and a 270-degree rotation. These rotations can be represented using matrix multiplication. Let's denote the no rotation as R0, the 90-degree rotation as R1, the 180-degree rotation as R2, and the 270-degree rotation as R3.
The matrix representation of these rotations can be written as follows:
R0 = [1 0]
[0 1]
R1 = [0 -1]
[1 0]
R2 = [-1 0]
[0 -1]
R3 = [0 1]
[-1 0]
Cyclic Group of Rotations:
If we consider the composition of rotations, we can see that rotating the square by 90 degrees four times brings it back to its original position. In other words, R1 * R1 * R1 * R1 = R4 = R0.
Similarly, rotating the square by 90 degrees three times gives us R3, rotating it by 90 degrees twice gives us R2, and rotating it by 90 degrees once gives us R1.
Therefore, if we consider the set {R0, R1, R2, R3}, we can see that it forms a cyclic group generated by R1. Every other rotation in the group can be expressed as a power of R1.
Hence, the set of four rotations of a square about a normal passing through the center of the plane forms a cyclic group.
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The set of four rotations of a square of its plane about a normal passing through the centre of the plane form aa)group but not abelianb)abelian group but not cyclicc)cyclic groupd)symmetric groupCorrect answer is option 'C'. Can you explain this answer?
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