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Math Help - 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
    G_1 is cyclic iff there exists a\in G_1 such that <br />
\left\langle a \right\rangle  = G_1<br />
. Okay, we now say, first, since there's a homomorphism: <br />
f\left( {a^n } \right) = f\left( a \right)f\left( {a^{n - 1} } \right) = ... = f^{n } \left( a \right)<br />
(1)

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

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

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

    Thus, for the first part:

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

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

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

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

    \Rightarrow\ x=f(a^n)=\left[f(a)\right]^n\in\left<f(a)\right>
    Last edited by JaneBennet; August 3rd 2008 at 10:06 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 \left<f(a)\right>\subseteq G_2 so we only need to show that G_2\subseteq \left<f(a)\right>.

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

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

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

    \Rightarrow\ x=f(a^n)=\left[f(a)\right]^n\in\left<f(a)\right>
    This shows that f:G_1\to G_2 only needs to be a surjective homomorphism in order for G_1 cyclic \implies 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 f:G_1\to G_2 only needs to be a surjective homomorphism in order for G_1 cyclic \implies G_2 cyclic to hold.
    Hence, if f is an isomorphism, the fact that f^{-1} is a surjective homomorphism from G_2 onto G_1 proves the reverse implication and completes the proof.
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