Are the groups $\mathbb C^\times \times \mathbb R^ \times$ and $\mathbb R^ \times\times \mathbb R^ \times$ isomorphic? Consider the groups $\mathbb R^ \times, \mathbb C^ \times$ under multiplication. I know that they are not isomorphic (as one of them is divisible but the other is not). My question is: 

Are the groups $\mathbb C^ \times\times \mathbb R^ \times$ and $\mathbb R^ \times \times \mathbb R^ \times$ isomorphic ? 

Please help. Thanks in advance.  
 A: It's possible to identify the isomorphism types of both groups in a fairly straightforward manner. Namely, thanks to the exponential map, we have an isomorphism
$$\mathbb{R}^{\times} \cong \mathbb{R} \times C_2$$
where $C_2$ denotes the cyclic group $\{ \pm 1 \}$ of order $2$, and similarly
$$\mathbb{C}^{\times} \cong \mathbb{R} \times S^1.$$
Abstractly, $\mathbb{R}$ is an uncountable-dimensional vector space over $\mathbb{Q}$, while $S^1$ is $\mathbb{R}/\mathbb{Z}$, hence is $\mathbb{Q}/\mathbb{Z}$ times another such thing. So we have decompositions
$$\mathbb{R}^{\times} \times \mathbb{R}^{\times} \cong C_2^2 \times \bigoplus_I \mathbb{Q}$$
where $I$ is uncountable, and 
$$\mathbb{C}^{\times} \times \mathbb{R}^{\times} \cong C_2 \times \mathbb{Q}/\mathbb{Z} \times \bigoplus_J \mathbb{Q}$$
where $J$ is uncountable. So both abelian groups are the direct sum of torsion and torsion-free groups, and have isomorphic torsion-free parts, but different torsion parts (namely $C_2^2$ and $C_2 \times \mathbb{Q}/\mathbb{Z}$). In particular, the second group has torsion of every possible order, but the first group only has torsion of order $1, 2$. 
A: If $\phi:G \rightarrow G'$ is an isomorphism, then an element $g \in G$ has order $n \iff \phi(g)$ has order $n$.   
Notice that $\mathbb{C}^\times \times \mathbb{R}^\times$ contains elements of order $4$, e.g. $(i, 1)$.  On the other hand, $\mathbb{R}^\times \times \mathbb{R}^\times$ does not contain elements of order $4$.
