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Es gibt genau einen Homomorphismus φ : Z12 → Z20 mit φ(1) = 15. Bestimmen Sie für diesen Homomorphismus ker(φ) und im(φ).

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1 Antwort

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Bei den kleinen Zahlen würde ich keine große Theorie anwenden, sondern einfach mit dem Rechnen anfangen:

\(\varphi([0]_{12}) = [0]_{20}\)

\(\varphi([1]_{12}) = [15]_{20}\)

\(\varphi([2]_{12}) = \varphi([1]_{12}+[1]_{12}) = \varphi([1]_{12}) + \varphi([1]_{12}) = [15]_{20}+ [15]_{20} = [15+15]_{20}=[30]_{20}=[10]_{20}\)

\(\varphi([3]_{12}) = \varphi([2]_{12}+[1]_{12}) = \varphi([2]_{12}) + \varphi([1]_{12}) = [10]_{20}+ [15]_{20} = [10+15]_{20}=[25]_{20}=[5]_{20}\)

\(\varphi([4]_{12}) = \varphi([3]_{12}+[1]_{12}) = \varphi([3]_{12}) + \varphi([1]_{12}) = [5]_{20}+ [15]_{20} = [5+15]_{20}=[20]_{20}=[0]_{20}\)

\(\varphi([5]_{12}) = \varphi([4]_{12}+[1]_{12}) = \varphi([4]_{12}) + \varphi([1]_{12}) = [0]_{20}+ [15]_{20} = [15]_{20}\)

\(\varphi([6]_{12}) = \varphi([4]_{12}+[2]_{12}) = \varphi([4]_{12}) + \varphi([2]_{12}) = [0]_{20}+ [10]_{20} = [10]_{20}\)

\(\varphi([7]_{12}) = \varphi([4]_{12}+[3]_{12}) = \varphi([4]_{12}) + \varphi([3]_{12}) = [0]_{20}+ [5]_{20} = [5]_{20}\)

\(\varphi([8]_{12}) = \varphi([4]_{12}+[4]_{12}) = \varphi([4]_{12}) + \varphi([4]_{12}) = [0]_{20}+ [0]_{20} = [0]_{20}\)

\(\varphi([9]_{12}) = \varphi([8]_{12}+[1]_{12}) = \varphi([8]_{12}) + \varphi([1]_{12}) = [0]_{20}+ [15]_{20} = [15]_{20}\)

\(\varphi([10]_{12}) = \varphi([8]_{12}+[2]_{12}) = \varphi([8]_{12}) + \varphi([2]_{12}) = [0]_{20}+ [10]_{20} = [10]_{20}\)

\(\varphi([11]_{12}) = \varphi([8]_{12}+[3]_{12}) = \varphi([8]_{12}) + \varphi([3]_{12}) = [0]_{20}+ [5]_{20} = [5]_{20}\)

Aus diesen Werten kann man dann ablesen ...

\(\text{ker}(\varphi) = \lbrace[0]_{12}, [4]_{12}, [8]_{12}\rbrace\)

\(\text{im}(\varphi) = \lbrace[0]_{20}, [5]_{20}, [10]_{20}, [15]_{20}\rbrace\)

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