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Suppose that phi : Z20--->Z20 is an automorphism and phi5=5.what are the various possibility for phi(x)?
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Suppose that phi : Z20--->Z20 is an automorphism and phi5=5.what are t...
Automorphisms in Z20


Introduction


In this question, we are given an automorphism, denoted as φ, from the set Z20 to itself. An automorphism is a bijective function that preserves the algebraic structure of the set. We are also given that φ(5) = 5. We need to determine the various possibilities for the function φ(x) and explain them in detail.

Solution


To find the possible values of φ(x), we need to consider the properties of automorphisms and the given information φ(5) = 5.

Properties of Automorphisms


An automorphism preserves the algebraic structure of the set. In the case of Z20, the algebraic structure includes addition and multiplication modulo 20. Therefore, for any elements a, b in Z20, the following properties hold for an automorphism φ:

1. φ(a + b) = φ(a) + φ(b) (Preservation of Addition)
2. φ(a * b) = φ(a) * φ(b) (Preservation of Multiplication)

φ(5) = 5


We are given that φ(5) = 5. This means that the automorphism φ preserves the element 5 in Z20. In other words, φ(5) is equal to 5. Using the preservation of addition property, we can write this as:

φ(5 + 0) = φ(5) + φ(0)
φ(5) = φ(5) + φ(0)
5 = 5 + φ(0)

From this equation, we can deduce that φ(0) must be equal to 0, as adding 0 to any number does not change its value.

φ(1)


Now, let's consider φ(1). We can express 1 as the sum of 5 and (-4) modulo 20:

1 = 5 + (-4) (mod 20)

Using the preservation of addition property, we can rewrite this equation as:

φ(1) = φ(5 + (-4))
φ(1) = φ(5) + φ(-4)
φ(1) = 5 + φ(-4)

To find the possible values of φ(1), we need to determine the possible values of φ(-4). Since φ(0) = 0, we can rewrite φ(-4) as φ(0 + (-4)):

φ(-4) = φ(0 + (-4))
φ(-4) = φ(0) + φ(-4)
φ(-4) = 0 + φ(-4)

From this equation, we deduce that φ(-4) can take any value in Z20.

Conclusion


In conclusion, the various possibilities for the function φ(x) are as follows:

1. φ(x) = x for all x in Z20 (Identity Function)
2. φ(x) = x + a for all x in Z20, where a is any constant in Z20 (Translation Function)
3. φ(x) = a for all x in Z20, where a is any constant in Z20 (Constant Function)

These possibilities arise from the preservation of addition and multiplication properties of autom
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Suppose that phi : Z20--->Z20 is an automorphism and phi5=5.what are the various possibility for phi(x)? for Mathematics 2025 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Suppose that phi : Z20--->Z20 is an automorphism and phi5=5.what are the various possibility for phi(x)? covers all topics & solutions for Mathematics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Suppose that phi : Z20--->Z20 is an automorphism and phi5=5.what are the various possibility for phi(x)?.
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