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Thread: iso and cyclic

  1. #1
    Junior Member mathemanyak's Avatar
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    iso and cyclic

    let f: G1->G2 be an isomorphism then show that G1 is cyclic iff G2 is cyclic
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  2. #2
    Super Member PaulRS's Avatar
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    Quote Originally Posted by mathemanyak View Post
    let f: G1->G2 be an isomorphism then show that G1 is cyclic iff G2 is cyclic
    $\displaystyle G_1$ is cyclic iff there exists $\displaystyle a\in G_1$ such that $\displaystyle
    \left\langle a \right\rangle = G_1
    $. Okay, we now say, first, since there's a homomorphism: $\displaystyle
    f\left( {a^n } \right) = f\left( a \right)f\left( {a^{n - 1} } \right) = ... = f^{n } \left( a \right)
    $ (1)

    Since $\displaystyle
    \left\langle a \right\rangle = G_1
    $ it must be because $\displaystyle
    G_1 = \left\{ {a,a^2 ,...,a^{\left| G_1 \right|} } \right\}
    $ (the order of a is |G_1|) being all those elements different. $\displaystyle f$ is biyective thus: $\displaystyle G_2=
    f\left( G_1 \right) = \left\{ {f\left( a \right),f\left( {a^2 } \right),...,f\left( {a^{\left| G_1 \right|} } \right)} \right\}
    $ by (1): $\displaystyle
    G_2 = \left\{ {f\left( a \right),f\left( {a^2 } \right),...,f\left( {a^{\left| {G_1 } \right|} } \right)} \right\} = \left\{ {f\left( a \right),f^2 \left( a \right),...,f^{^{\left| {G_1 } \right|} } \left( a \right)} \right\}
    $ thus $\displaystyle
    G_2 = \left\langle {f\left( a \right)} \right\rangle
    $

    The converse is analogous, take $\displaystyle
    f^{ - 1} :G_2 \to G_1
    $
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  3. #3
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by PaulRS View Post
    $\displaystyle G_1 = \left\{ {a,a^2 ,...,a^{\left| G_1 \right|} } \right\}$
    You are assuming that $\displaystyle G_1$ is finite. The question doesn’t say anything about finite groups, though.

    If $\displaystyle G_1=\left<a\right>$ is cyclic and $\displaystyle f:G_1\to G_2$ is an isomorphism, you just have to show that $\displaystyle G_2=\left<f(a)\right>$. Similarly, if $\displaystyle G_2=\left<b\right>$ is cyclic, then $\displaystyle G_1=\left<f^{-1}(b)\right>$.

    Thus, for the first part:

    We have that $\displaystyle \left<f(a)\right>\subseteq G_2$ so we only need to show that $\displaystyle G_2\subseteq \left<f(a)\right>$.

    $\displaystyle \color{white}.\quad.$ $\displaystyle x\in G_2$

    $\displaystyle \Rightarrow\ f^{-1}(x)\in G_1=\left<a\right>$

    $\displaystyle \Rightarrow\ f^{-1}(x)=a^n$ for some $\displaystyle n\in\mathbb{Z}$

    $\displaystyle \Rightarrow\ x=f(a^n)=\left[f(a)\right]^n\in\left<f(a)\right>$
    Last edited by JaneBennet; Aug 3rd 2008 at 09:06 AM.
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  4. #4
    MHF Contributor

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    JaneBennet is right but, considering the fact that PaulRS is only 18 years old, it's fascinating that how knowledgeable he is!
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  5. #5
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    Quote Originally Posted by JaneBennet View Post
    Thus, for the first part:

    We have that $\displaystyle \left<f(a)\right>\subseteq G_2$ so we only need to show that $\displaystyle G_2\subseteq \left<f(a)\right>$.

    $\displaystyle \color{white}.\quad.$ $\displaystyle x\in G_2$

    $\displaystyle \Rightarrow\ f^{-1}(x)\in G_1=\left<a\right>$

    $\displaystyle \Rightarrow\ f^{-1}(x)=a^n$ for some $\displaystyle n\in\mathbb{Z}$

    $\displaystyle \Rightarrow\ x=f(a^n)=\left[f(a)\right]^n\in\left<f(a)\right>$
    This shows that $\displaystyle f:G_1\to G_2$ only needs to be a surjective homomorphism in order for $\displaystyle G_1$ cyclic $\displaystyle \implies$ $\displaystyle G_2$ cyclic to hold.
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  6. #6
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by algebraic topology View Post
    This shows that $\displaystyle f:G_1\to G_2$ only needs to be a surjective homomorphism in order for $\displaystyle G_1$ cyclic $\displaystyle \implies$ $\displaystyle G_2$ cyclic to hold.
    Hence, if $\displaystyle f$ is an isomorphism, the fact that $\displaystyle f^{-1}$ is a surjective homomorphism from $\displaystyle G_2$ onto $\displaystyle G_1$ proves the reverse implication and completes the proof.
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