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Thread: group homomorphism

  1. #1
    Member Mauritzvdworm's Avatar
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    group homomorphism

    Let $\displaystyle G\text{ and }H$ be additive subgroups of $\displaystyle \mathbb{R}$ both containing 1. Let $\displaystyle \eta:G\rightarrow H$ be an order preserving group homomorphism such that $\displaystyle \eta(1)=1$. Show that $\displaystyle \eta(g)=g$ for all $\displaystyle g\in G$.

    The idea that I am working with is to let $\displaystyle s,t\in G$ such that
    $\displaystyle s-t>1$ then there exists an integer $\displaystyle m$ such that
    $\displaystyle s>m>t$.

    now apply $\displaystyle \eta$ and get the following
    $\displaystyle \eta(s)-\eta(t)>1$ and $\displaystyle \eta(s)>\eta(m)>\eta(t)$
    and suppose that $\displaystyle \eta(g)>g$ for all $\displaystyle g\in G$. In this way I hope to get some contradiction

    Or is there another way?
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  2. #2
    Member Mauritzvdworm's Avatar
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    This problem is still bugging me... does anybody have any ideas?
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