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Math Help - please check if my proof is correct

  1. #1
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    please check if my proof is correct

    Check if \langle t^{n} \rangle is a retract of \langle t \rangle.

    Proof:
    Define \phi : \langle t \rangle \rightarrow \langle t \rangle by \phi (x)=x for all x \in \langle t \rangle.

    Let x_{1}, x_{2} \in \langle t \rangle.
    Then x_{1}=t^{a}, x_{2}=t^{b} where a,b are integers.

    \phi (x_{1}x_{2})= \phi (t^{a}t^{b})=\phi (t^{a+b})=t^{a+b}=t^{a}t^{b}=\phi (t^{a} \phi (t^{b})=\phi (x_{1}) \phi(x_{2})
    So, \phi is homomorphism.

    Clearly, \phi (x^*)=x^* for all x^* \in \langle t^{n} \rangle.

    Since \phi (x_{1})= \phi (t^{a})=t^{a}, t^{a} \in \langle t^{n} \rangle \Leftrightarrow a | n.

    Hence, \langle t^{n} \rangle need not be a retract of \langle t \rangle.
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  2. #2
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    Quote Originally Posted by deniselim17 View Post
    Check if \langle t^{n} \rangle is a retract of \langle t \rangle.

    Proof:
    Define \phi : \langle t \rangle \rightarrow \langle t \rangle by \phi (x)=x for all x \in \langle t \rangle.

    Let x_{1}, x_{2} \in \langle t \rangle.
    Then x_{1}=t^{a}, x_{2}=t^{b} where a,b are integers.

    \phi (x_{1}x_{2})= \phi (t^{a}t^{b})=\phi (t^{a+b})=t^{a+b}=t^{a}t^{b}=\phi (t^{a} \phi (t^{b})=\phi (x_{1}) \phi(x_{2})
    So, \phi is homomorphism.

    Clearly, \phi (x^*)=x^* for all x^* \in \langle t^{n} \rangle.

    Since \phi (x_{1})= \phi (t^{a})=t^{a}, t^{a} \in \langle t^{n} \rangle \Leftrightarrow a | n.

    Hence, \langle t^{n} \rangle need not be a retract of \langle t \rangle.

    What is t, anyway? A normal subgroup of a group is a retract of the group iff it has a normal complement.

    Tonio
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  3. #3
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    \langle t \rangle is a cyclic group generated by t.
    t is a generator.
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  4. #4
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    Quote Originally Posted by deniselim17 View Post
    \langle t \rangle is a cyclic group generated by t.
    t is a generator.

    Good. So <t> is a cyclic group and thus it cannot be expressed as a non-trivial direct product ==> no subgroup of <t> has normal complement ==> there exists no retract of <t>.

    Tonio
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