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    $(\RR,+)\cong(\RR_{>0},\times)$, but not for $\QQ$

    P-R5BKY

    Area
    Algebra
    Topics
    Abelian Groups, Counterexamples, Homomorphisms
    Status
    draft

    Is \((\RR,+)\) isomorphic to \((\RR_{>0},\times)\)? Is the analogous statement true for \(\QQ\) and \(\QQ_{>0}\)?

    Solution

    For the reals, yes. The exponential map \[ \exp:(\RR,+)\longrightarrow(\RR_{>0},\times) \] is a group isomorphism, with inverse \(\log\).

    For the rationals, no. The additive group \((\QQ,+)\) is divisible: for every \(q\in\QQ\) and every positive integer \(n\), there exists \(x=q/n\) with \[ nx=q. \]

    The multiplicative group \((\QQ_{>0},\times)\) is not divisible. For example, \(2\) has no square root in \(\QQ_{>0}\). Divisibility is preserved by group isomorphism, so \[ (\QQ,+)\not\cong(\QQ_{>0},\times). \]

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