TITLE: Multiplicative groups
QUESTION [6 upvotes]: Consider the multiplicative groups $\def\R{{\mathbb R}}(\R^\ast,\cdot)$ and $(\R^+, \cdot)$.  Show that $\R^\ast \cong\R^+\times\Bbb Z_2$.
Consider the  multiplicative group $\def\Q{{\mathbb Q}}(\Q^+,\cdot)$. List the elements of the cyclic subgroup $H=\langle4\rangle$. Is $H$ normal?  List the elements of the cosets $\frac13H$ and $2H$. Describe the quotient $\Q^+/H$ by picking a representative for each coset of $H$.  In the quotient $\Q^+/H$, what is the order of $\frac13H$? What is the order of $2H$?  What is $(2H)\Bigl(\frac23H\Bigr)$?


For $(1)$, I thought if $a$ is in $\mathbb{R}^*$, then we can find function $F$ such that $F(a) = a \times 1$ if $a>0$, 
and $F(a) = -a \times 0$ if $a<0$. Maybe this function shows they are isomorphic? Is this right?  
For $(2)$, how $H=\langle 4\rangle$ is cyclic subgroup? Is $H$ all $\mathbb{Q}*$, since if $a$ is in $\mathbb{Q}*$, we can always find $\frac14a$ in $\mathbb{Q}*$ or is $H$ just $\{4,8,12,16,\dots\}$ or $\{\dots,-8,-4,0,4,8.12,\dots\}$ ? I'm not sure what $H$ is.
To describe the quotient $Q^+$/H, maybe it is a set of aH such that 
$0 < a\leq 1$ and 
$a= \dfrac{x}{y}$ such that x and y are integers and $gcd(x,y) = 1$. 
Thank you!
(Original image of questions: http://i.stack.imgur.com/wRRUJ.png)

REPLY [2 votes]: For $(1)$, all you need to do is check the map you defined is in fact an isomorphism (check it's a homomorphism, check its image, and check its kernel)
In response to $(2)$, $H$ is the group of elements {$ 4^n| n\epsilon \mathbb Z $}. The multiplicative group $(\mathbb Q^+,*)$ is the group of the positive rational numbers under multiplacation in the usual way. H is normal since every subgroup of an abelian group is normal. You can represent each of the cosets with $m/n$ where $4$ does not divide $n$ or $m$. This representation is obviously unique and any two cosets have different representations of this form. Some elements in $G/H$ are of finite order some are not. This should be enough to do the problem.