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?
Best Answer
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.