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Math Help - Mapping between two cyclic groups

  1. #1
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    Mapping between two cyclic groups

    Let G be a cyclic group of order 5. Let G' be a cyclic group of order 7. Let f : G -> G' be a homomorphism. Why is f the trivial homomorphism?

    In my group, the trivial homomorphism means the map that takes every element of the domain and sends it to the e element of the codomain.

    I know what G and G' look like.

    G = \{e, a, a^2, a^3, a^4\} and G' = \{e, b, b^2, b^3, b^4, b^5, b^6\}

    But how is it that f must be the trivial homomorphism?
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  2. #2
    Senior Member roninpro's Avatar
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    Try using the fact that a homomorphism takes subgroups to subgroups. What are the subgroups of the cyclic group of order 7?
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  3. #3
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    I found this theorem in my book. Let f : G -> G' be a group-homomorphism. The image of f is a subgroup of G'.

    So by this theorem, f(G) is a subgroup of G'.

    Then I found this online: Let a cyclic group have order n. If r is a divisor of n then there is some number s such that rs = n. Let a be an element of the group with order n. (it must have one if it is cyclic) Then ar is of order s and therefore generates a cyclic subgroup of order s while as generates a subgroup of order r.

    So if that is true, since 7 is a prime the only subgroups G' has is of order 1 or 7. This would imply that f is the trivial homomorphism.

    Is this correct?
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  4. #4
    Senior Member roninpro's Avatar
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    That's right. You have to send G to G' or \{e'\}. The former isn't possible because it is too large, so G\mapsto \{e'\}, which is the trivial homomorphism.

    Nicely done.
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