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Is the group $\mathbb Q^*$ (rationals without $0$ under multiplication) a direct product or a direct sum of nontrivial subgroups?

My thoughts:

Consider subgroups $\langle p\rangle=\{p^k\mid k\in \mathbb Z\}$ generated by a positive prime $p$ and $\langle -1\rangle=\{-1,1\}$.

They are normal (because $\mathbb Q^*$ is abelian), intersects in $\{1\}$ and any $q\in \mathbb Q^*$ is uniquely written as quotient of primes' powers (finitely many).

So, I think $\mathbb Q^*\cong \langle -1\rangle\times \bigoplus_p\langle p\rangle\,$ where $\bigoplus$ is the direct sum.

And simply we can write $\mathbb Q^*\cong \Bbb Z_2\times \bigoplus_{i=1}^\infty \Bbb Z$.

Am I right?

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    $\begingroup$ Of course, each $\langle p\rangle\cong\mathbb Z$, so we might write this isomorphism using familiar groups: $$\mathbb Q^*\cong\mathbb Z_2\times\bigoplus_{i\in\mathbb N}\mathbb Z$$ $\endgroup$ – user714630 May 10 '13 at 13:56
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Yes, you're right.

Your statement can be generalized to the multiplicative group $K^*$ of the fraction field $K$ of a unique factorization domain $R$. Can you see how?

In fact, if I'm not mistaken it follows from this that for any number field $K$, the group $K^*$ is the product of a finite cyclic group (the group of roots of unity in $K$) with a free abelian group of countable rank, so of the form

$K^* \cong \newcommand{\Z}{\mathbb{Z}}$ $\Z/n\Z \oplus \bigoplus_{i=1}^{\infty} \Z.$

Here it is not enough to take the most obvious choice of $R$, namely the full ring of integers in $K$, because this might not be a UFD. But one can always choose an $S$-integer ring (obtained from $R$ by inverting finitely many prime ideals) with this property and then apply Dirichlet's S-Unit Theorem.

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