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Thread: Homomorphism and periods

  1. #1
    Junior Member
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    Homomorphism and periods

    Let $\displaystyle f : G -> G'$ be a homomorphism. Let $\displaystyle a$ be an element of $\displaystyle G$ of period 10. Prove that $\displaystyle f(a)$ must have period 1 or 2 or 5 or 10.

    I can prove that $\displaystyle f(a)$ may have period 10.

    $\displaystyle a^{10} = e$ because $\displaystyle a$ have period 10
    $\displaystyle f(a^{10}) = [f(a)]^{10} = e'$ by property of homomorphism
    Therefore $\displaystyle f(a)$ may have period 10.
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  2. #2
    Senior Member roninpro's Avatar
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    Suppose that $\displaystyle f(a)$ has period $\displaystyle k$. Then by the division algorithm, we can write $\displaystyle 10=qk+r$ where $\displaystyle q,r$ are integers and $\displaystyle 0\leq r<k-1$. Then, following your original calculation,

    $\displaystyle e'=f(a^{10})=f(a^{qk+r})=f(a^{qk})f(a^r)=[f(a)]^{qk}[f(a)]^r$

    First note that $\displaystyle [f(a)]^{qk}=e'$ as $\displaystyle k$ was the period. This means that $\displaystyle [f(a)]^r=e'$. But since $\displaystyle r<k$, we must take $\displaystyle r=0$ (i.e. $\displaystyle k$ was the smallest nonzero number such that $\displaystyle [f(a)]^k=e'$). This means that $\displaystyle 10=qk$, so the period $\displaystyle k$ must divide 10. So we can conclude that $\displaystyle k=1, 2, 5,\text{ or } 10$.
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