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An exercise asks me to show that $\mathbb Q \times C_2$ (under addition of the rationals) is isomorphic to $\mathbb Q-\{0\}$ under multiplication. Quotienting $\mathbb Q \times C_2$ by $C_2$ gives $\mathbb Q$ under addition. The only element of order 2 in $\mathbb Q-\{0\}$ is -1, so quotienting out by $<-1>$ gives $\mathbb Q^+$ under multiplication. But I don't think $\mathbb Q$ under addition and $\mathbb Q^+$ under multiplication can be isomorphic. If $f$ were such an isomorphism from the additive group to the multiplicative group, then $f\left(\frac{a}{b}\right)$ would have to be $f(1)^\frac{a}{b}$. But no matter which positive rational $r$ I pick to be $f(1)$ there will be a rational $q$ such that $r^q$ is irrational, unless I let $r=1$ in which case $f$ would not be injective.

Where did I mess up? Am I not allowed to say that the quotients of both groups by $C_2$ are isomorphic?

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  • $\begingroup$ what's $C_2$ ? (obligatory extra characters) $\endgroup$
    – Stefan Smith
    Commented Dec 31, 2013 at 1:10
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    $\begingroup$ The cyclic group of order $2$, I assume. $\endgroup$
    – Dylan Yott
    Commented Dec 31, 2013 at 1:10
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    $\begingroup$ Relevant: math.stackexchange.com/questions/33607/… $\endgroup$
    – Dylan Yott
    Commented Dec 31, 2013 at 1:11
  • $\begingroup$ Yes, it's the cyclic group of order 2. So did the exercise ask me to prove a false statement? $\endgroup$
    – Nishant
    Commented Dec 31, 2013 at 1:19
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    $\begingroup$ $\mathbb{Q}^{\ast}$ should be isomorphic to $ \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$, not $ \mathbb{Z}/2\mathbb{Z} \times \mathbb{Q}$. The isomorphism is given by $$(-1)^{\epsilon} \prod\limits_p p^{a_p} \mapsto ((\epsilon + 2 \mathbb{Z}), (a_p)_p)$$ where $p$ runs through the prime numbers. This follows from the fundamental theorem of arithmetic. $\endgroup$
    – D_S
    Commented Jun 8, 2017 at 13:24

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