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Let be a finite group of order and let be any integer relatively prime to . Define : → by () = ^.

(a) Prove that is one-to-one.

User CLearner
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Final answer:

To prove that the function ϕ is one-to-one, we need to show that for every pair of elements € and €', if ϕ(€) = ϕ(€'), then € = €'. This can be done by assuming that ϕ(€) = ϕ(€') and deriving a contradiction. By showing that the only way for the equation to hold is if € = €', we prove that ϕ is one-to-one.

Step-by-step explanation:

In order to prove that the function ϕ is one-to-one, we need to show that for every pair of elements € and €', if ϕ(€) = ϕ(€'), then € = €'.

To do this, let's assume that ϕ(€) = ϕ(€'). Using the definition of ϕ given in the question, this means that €^€' = €'^€. Taking the logarithm of both sides, we get log€^€' = log€'^€. Using the property of logarithms, we can rewrite this as €'log€ = €log€'. Since € and €' are relatively prime to , the only way for this equation to hold is if € = €', thus proving that ϕ is one-to-one.

User Lenny Carmi
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