Hello,
Please check if my answers are corrects :
1- What are the elements of order 3 in the additive group Z/3Z × Z/6Z ?
To find the elements of order 3, we need to look for elements whose multiplication by 3 is equivalent to the neutral element (0, 0). This means we are looking for elements (a, b) such that 3(a, b) = (0, 0) in the group.
Therefore, the elements of order 3 in the group Z/3Z × Z/6Z are:
(0, 0),(1, 0), (2, 0), (0, 1), (1, 1), (2, 1), (0, 2), (1, 2), (2, 2).(2, 2),(0, 4),(1, 4),(2, 4)
2- Is there a surjective group homomorphism ((Q,+) → (Q^+\{0},×)?
Let's consider the exponential function exponentiation(x) = e^x, where e is the base of the natural logarithm. This function is well-defined and continuous for all real numbers x, including rational numbers. Moreover, it is strictly increasing over its entire domain.
Now, let's define a function f: Q → Q^+\{0} as follows:
f(x) = e^x
We can verify that this function satisfies the required properties to be a group homomorphism:
1. The function f preserves the group operation. For any rational numbers x and y, we have f(x + y) = e^(x + y) = e^x * e^y = f(x) * f(y). This means that addition in the source group (Q, +) is preserved in the target group (Q^+\{0}, ×).
2. The function f is surjective. Since the exponential function is strictly increasing over its domain, it will take on all positive non-zero values in the image. Therefore, for any positive non-zero rational number q in the target group (Q^+\{0}, ×), there exists a rational number x in the source group (Q, +) such that f(x) = q.
By combining these two properties, we can conclude that the function f: Q → Q^+\{0} defined by f(x) = e^x is a surjective group homomorphism from the additive group of rational numbers (Q, +) to the multiplicative group of positive non-zero rational numbers (Q^+\{0}, ×).
- Does the symmetric group S10 contains elements of order 30 ?
The symmetric group S10 does indeed contain elements of order 30.
An example of such an element is the following permutation:
(1 2 3 4 5 6 7 8 9 10)(2 3)(4 5)(6 7)(8 9)(1 10)
This permutation cycles through the elements from 1 to 10 and exchanges the pairs of elements: (2, 3), (4, 5), (6, 7), (8, 9), and (1, 10). Applying this permutation 30 times yields the identity element of S10, indicating that its order is indeed 30.
There are several other permutations of order 30 in the symmetric group S10. This example is just one among many. Permutations of order 30 can be constructed using different cycles and exchanges within S10.
Please check if my answers are corrects :
1- What are the elements of order 3 in the additive group Z/3Z × Z/6Z ?
To find the elements of order 3, we need to look for elements whose multiplication by 3 is equivalent to the neutral element (0, 0). This means we are looking for elements (a, b) such that 3(a, b) = (0, 0) in the group.
Therefore, the elements of order 3 in the group Z/3Z × Z/6Z are:
(0, 0),(1, 0), (2, 0), (0, 1), (1, 1), (2, 1), (0, 2), (1, 2), (2, 2).(2, 2),(0, 4),(1, 4),(2, 4)
2- Is there a surjective group homomorphism ((Q,+) → (Q^+\{0},×)?
Let's consider the exponential function exponentiation(x) = e^x, where e is the base of the natural logarithm. This function is well-defined and continuous for all real numbers x, including rational numbers. Moreover, it is strictly increasing over its entire domain.
Now, let's define a function f: Q → Q^+\{0} as follows:
f(x) = e^x
We can verify that this function satisfies the required properties to be a group homomorphism:
1. The function f preserves the group operation. For any rational numbers x and y, we have f(x + y) = e^(x + y) = e^x * e^y = f(x) * f(y). This means that addition in the source group (Q, +) is preserved in the target group (Q^+\{0}, ×).
2. The function f is surjective. Since the exponential function is strictly increasing over its domain, it will take on all positive non-zero values in the image. Therefore, for any positive non-zero rational number q in the target group (Q^+\{0}, ×), there exists a rational number x in the source group (Q, +) such that f(x) = q.
By combining these two properties, we can conclude that the function f: Q → Q^+\{0} defined by f(x) = e^x is a surjective group homomorphism from the additive group of rational numbers (Q, +) to the multiplicative group of positive non-zero rational numbers (Q^+\{0}, ×).
- Does the symmetric group S10 contains elements of order 30 ?
The symmetric group S10 does indeed contain elements of order 30.
An example of such an element is the following permutation:
(1 2 3 4 5 6 7 8 9 10)(2 3)(4 5)(6 7)(8 9)(1 10)
This permutation cycles through the elements from 1 to 10 and exchanges the pairs of elements: (2, 3), (4, 5), (6, 7), (8, 9), and (1, 10). Applying this permutation 30 times yields the identity element of S10, indicating that its order is indeed 30.
There are several other permutations of order 30 in the symmetric group S10. This example is just one among many. Permutations of order 30 can be constructed using different cycles and exchanges within S10.
