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Math Help - phi: A --> G is a group homomorphism. Prove that a acts by the rule a.x=phi(a).x

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    phi: A --> G is a group homomorphism. Prove that a acts by the rule a.x=phi(a).x

    Suppose that G is a group acting on X and that  \phi : A \rightarrow G is a group homomorphism. Prove that a acts on X by the rule a.x = \phi (a).x.
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    Quote Originally Posted by joestevens View Post
    Suppose that G is a group acting on X and that \phi : A \rightarrow G is a group homomorphism. Prove that a acts on X by the rule a.x = \phi (a).x.
    It pretty much falls out if you just follow the definition. What are you having trouble with?
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    Definition: G acts on X if there exists a function \alpha : G \times X \rightarrow X such that
    i) for g,h \in G,\ \alpha_g \circ \alpha_h = \alpha_{gh}
    ii) \alpha_e = e_X.

    Should I use i & ii above as the definition of the group action G on X and show A also acts on X by a.x = \phi(a).x?

    Am I to take A as a subgroup of G or is A just any group?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by joestevens View Post
    Definition: G acts on X if there exists a function \alpha : G \times X \rightarrow X such that
    i) for g,h \in G,\ \alpha_g \circ \alpha_h = \alpha_{gh}
    ii) \alpha_e = e_X.

    Should I use i & ii above as the definition of the group action G on X and show A also acts on X by a.x = \phi(a).x?

    Am I to take A as a subgroup of G or is A just any group?
    Just use that definition. For example, i) is satisfied since (ab)\cdot x=\phi(ab)x=\phi(a)\phi(b)x=a\cdot(\phi(b)x)=a\cdo  t(b\cdot x)
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