Recently I am reading Nathan Carter's Visual Group Theory. I saw this exercise:

Exercise 8.40. Recall the group $\mathbb Q$ (under addition) and the group $\mathbb Q^*$ (under multiplication) introduced in Exercise 4.33. Show that $\mathbb Q\times C_2 \cong \mathbb Q^*$ by specifying the isomorphism, and explaining why the function you give is indeed an isomorphism.

Here $\mathbb Q^*$ is the set of non-zero rational numbers and $C_2$ is the cyclic group of order 2.

But I think that I am not able to construct the isomorphism.
If the $\mathbb Q$ is replaced by $\mathbb R$ instead, then it is much easier:
Denote $C_2$ as $(\{1,-1\},\times)$,
one of the isomorphism is $f(a,b)=be^a$.

But now we are dealing with $\mathbb Q$, and most number to a rational number power is not rational (whatever base chosen).
So here's my question: am I on the right track? Is this done by exponential and I have missed something? Or it is nothing about exponential but actually something else?

Thanks in advance!

  • 2
    $\begingroup$ It can't even be possible to have a homomorphism $\mathbb Q\to\mathbb Q^{*}$ other than the trivial one. If $x$ is in the image of such a homomorphism, then so is $x^{1/n}$ for each integer $n\neq 0$... $\endgroup$ – Thomas Andrews Nov 20 '16 at 3:57
  • $\begingroup$ What is $e$ in your definition of $f$? $\endgroup$ – Sayantan Nov 20 '16 at 4:33
  • $\begingroup$ @ThomasAndrews I see...could I justify it by saying that if $p^n|x$, $p$ is prime and $n$ is largest possible, then $x^{1/(n+1)}$ is not rational? $\endgroup$ – Brian Cheung Nov 20 '16 at 6:44
  • $\begingroup$ @Sayantan It is the Euler's number, not the identity element in group. $\endgroup$ – Brian Cheung Nov 20 '16 at 6:45

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