finding a map s.t. $\mathbb Q \times C_2$ $\cong$ $\mathbb Q^*$ This is a question from my group theory exam which I was unable to prove:
It say that Show that $\mathbb Q \times C_2$ $\cong$ $\mathbb Q^*$ by specifying an isomorphism.
But I couldn't find one. How to find such a map?
(Here $\mathbb Q$ is considered a group under addition and $\mathbb Q^*$ is considered a group under multiplication.)
 A: There is only one element of order $2$ in $\mathbb{Q}^*$, namely $-1$. The only element of order $2$ in $\mathbb{Q}\times C_2$ is $(0,1)$ (I'll write also $C_2=\{0,1\}$ additively). So, if $f\colon\mathbb{Q}\times C_2\to\mathbb{Q}^*$ is an isomorphism, the subgroup $H=\{(0,0),(0,1)\}$ of the domain is mapped onto $K=\{1,-1\}$, which means that $f$ induces an isomorphism
$$
g\colon\frac{\mathbb{Q}\times C_2}{H}\to\frac{\mathbb{Q}^*}{K}
$$
However, $\mathbb{Q}\times C_2/H\cong\mathbb{Q}$ is divisible, while $\mathbb{Q}^*/K$ is isomorphic to the group of positive rationals under multiplication which is not divisible, because it doesn't contain $\sqrt{2}$.
This is a contradiction, so $f$ doesn't exist in the first place.

What's a divisible group? An additive group $G,+$ is divisible when, for all $n>0$ and for all $g\in G$, there exists $h\in G$ such that $nh=g$. If the group is multiplicative the condition reads $h^n=g$ (just due to the different notation). Examples of divisible group are $\mathbb{Q}$, $\mathbb{R}_{>0}$ (the multiplicative group of positive reals) and $\mathbb{C}^*$ (the multiplicative group of nonzero complex numbers). Every quotient of a divisible group is divisible.
A: You are to be congratulated for not finding an isomorphism, as these groups are not isomorphic.  The multiplicative group $\mathbb{Q}^{\ast}$ is isomorphic to $C_2\times\mathbb{Z}^{\omega}$, a direct sum of cyclic groups, while the additive group $\mathbb{Q}$ is divisible.
Added: If, according to @Timbuc's suggestion, it was $\mathbb{R}$ that was meant, then there is indeed an isomorphism.  If we think of $C_2$ as the group $\{1,-1\}$ under multiplication, then an isomorphism $\mathbb{R}\times C_2 \to \mathbb{R}^{\ast}$ is given by $(r,\sigma)\mapsto \sigma e^r$.
