Prove that $\mathbb{Q}^{\times}$ not isomorphic to $\mathbb{Z}^{n}$ $\mathbb{Q}^{\times}$ is the group of rational number without $0$ under multiplication, and $\mathbb{Z}^{n}$ is the free abelian group of rank $n$. Show that $\mathbb{Q}^{\times}$ not isomorphic to $\mathbb{Z}^{n}$. I tried to assume that there is an isomorphism and to get a contradiction.. didn't succeed
 A: $\mathbf{Q}^{\times}$ is generated by the set of positive prime numbers and $-1$ so it contains (a groupe isomorphic to) $\mathbf{Z}^{\mathscr{P}}$ where $\mathscr{P}$ is the set of positive prime numbers. $\mathbf{Z}^{\mathscr{P}}$ is not of finite rank over $\mathbf{Z}$ whereas any $\mathbf{Z}^n$ is, so there's no possible isomorphism $\mathbf{Q}^{\times} \rightarrow \mathbf{Z}^n$, for any $n$, as any such arrow would give an infinite free family in $\mathbf{Z}^n$, namely, the family image of $\mathscr{P}$ by the supposed isomorphism.
To put if in a "fashion" way the morphism $$v : \mathbf{Q}^{\times} \rightarrow \{\pm\}\times \mathbf{Z}^{\mathscr{P}}$$ defined by $v(x) = \left( \textrm{sign}(x), \left(   v_p(x)\right)_{p\in \mathscr{P}} \right)$ where $v_p$ is the $p$-adic valuation function for $p\in \mathscr{P}$, is an isomorphism.
This relies on the fact that $\mathscr{P}$ is infinite - fact admitted here. ;-)
A: Suppose you had a homomorphism $f:\mathbb Q^{\times}\rightarrow\mathbb Z^{n}$. Notice that $(-1)^2=1$ thus $f(-1)^2=1$. The only solution to $x^2=e$ (where $e$ is the identity) in $\mathbb Z^{n}$ is $e$, thus $f(-1)=f(1)=e$, meaning no homomorphism is injective and the groups are not isomorphic.
(To be clear, notice that $x^2=e$ in $\mathbb Z^n$ would more commonly be written as $x+x=0$, but I chose to write it multiplicatively to aid seeing the connection between $(-1)^2=1$ (in $\mathbb Q^{\times}$) and $x^2=e$ (in $\mathbb Z^{n}$))
