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Question: Let $G=(\mathbb{R}^{*},\cdot)$, be the multiplicative group of nonzero real numbers. Recall that $\mathbb{Z}_{2}=\{0,1\}$ with the operation $0+0=0$, $0+1=1$, $1+0=0$, $1+1=0$ is a group. If we let $\mathbb{R}$ represent the group of all reals with addition, find a homomorphism
$\phi :\mathbb{R}\times \mathbb{Z}_{2} \rightarrow G$
that is also a bijection.

What I know so far: A homomorphism is injective iff its kernel K is the trivial subgroup {1} of G. Therefore, I'll need the identity elements of $\mathbb{R}\times \mathbb{Z}_{2}$ to be mapped to the identity of G.

So, $(0,0)\rightarrow (1)$.

Now, I should find maps such that:
$(0,1)\rightarrow (?)$
$(r,0)\rightarrow (?)$
$(R,1)\rightarrow (?)$

If I'm on the right track, I'm not sure how to structure the maps such that $\phi $ is 1-1 and onto.

If I'm not on the right track, some advice would be appreciated. Thanks!

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it would be better to think of ${\Bbb{Z}}_2$ as $\{1,-1\}$ with multiplication – janmarqz Feb 5 '14 at 4:33

Suppose $\phi:\ \Bbb{R}\times\Bbb{Z}_2\ \longrightarrow\ \Bbb{R}^*$ is an isomorphism (i.e. a bijective homomorphism). As you noted it satisfies $\phi((0,0))=1$. Moreover the number $\phi((0,1))\in\Bbb{R}^*$ satisfies $$(\phi((0,1)))^2=\phi((0,1))\cdot\phi((0,1))=\phi((0,1)+(0,1))=\phi((0+0,1+1))=\phi((0,0))=1,$$ which shows that $\phi((0,1))=\pm1$. Because $\phi$ is injective we cannot have $\phi((0,1))=1$, and so $\phi((0,1))=-1$. It follows that for all $(x,y)\in\Bbb{R}\times\Bbb{Z}_2$ we have $$\phi((x,y))=\phi((x,0)+(0,y))=\phi((x,0))\cdot\phi((0,y))=(-1)^y\cdot\phi((x,0)),$$ which explains janmarqz's comment. It remains to figure out what values $\phi$ takes on $\Bbb{R}\times\{0\}$, so consider the map $$\phi':\ \Bbb{R}\ \longrightarrow\ \Bbb{R}^*:\ x\ \longmapsto\ \phi((x,0)).$$ It is an injective group homomorphism. Constructing a map $\phi'$ with the required properties only from the given data is not easy; there are infinitely many such maps. You have probably seen an example of a map $\phi':\ \Bbb{R}\ \longrightarrow\ \Bbb{R}^*$ satisfying $\phi'(x+y)=\phi'(x)\cdot\phi'(y)$ and $\phi'(0)=1$ already. Can you think of one?

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