# Group $\mathbb Q^*$ as direct product/sum

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?

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

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.