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  1. #1
    mms
    mms is offline
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    Groups

    Show that if <br />
f:G \to H<br /> <br />
is a homomorphism, <br />
a \in G<br /> <br />
    and <br />
f(a)<br /> <br />
has a finite order in H then
    <br />
\left| a \right|<br /> <br />
is infinite or <br />
\left| {f(a)} \right|<br /> <br />
divides <br />
\left| a \right|<br /> <br />
    thank you.
    Last edited by mms; July 18th 2009 at 08:24 PM.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by mms View Post
    Show that if <br />
f:G \to H<br /> <br />
is a homomorphism, <br />
a \in G<br /> <br />
    and <br />
f(a)<br /> <br />
has a finite order in H then
    <br />
\left| a \right|<br /> <br />
is infinite or <br />
\left| {f(a)} \right|<br /> <br />
divides <br />
\left| a \right|<br /> <br />
    thank you.
    if |a|=n < \infty, then 1=f(1)=f(a^n)=(f(a))^n and thus |f(a)| \mid n.
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  3. #3
    mms
    mms is offline
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    thank you, but i have a doubt, if f(a) has a finite order does that mean that a has a finite order?
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  4. #4
    MHF Contributor

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    Quote Originally Posted by mms View Post
    thank you, but i have a doubt, if f(a) has a finite order does that mean that a has a finite order?
    not necessarily, but if the order of a is finite, then it's divisible by the order of f(a). (see your original question again!)

    example: define f: \mathbb{Z} \longrightarrow \mathbb{Z} by f(k)=0, for all k \in \mathbb{Z}. then f is a homomorphism. let a=1. then |f(a)|=|0|=1 but |a| is not finite.
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