# $\mathbb{R} \setminus \{ 0 \}$ direct product

Is it possible to express $$\mathbb{R} \setminus \{ 0 \}$$ as a direct product of $$H$$ and $$K$$ where $$H$$ and $$K$$ are subgroups of $$\mathbb{R} \setminus \{ 0 \}$$. I know that $$\mathbb{R} \setminus \{ 0 \} \cong \mathbb{R} \times \mathbb{Z}_2$$. But here $$\mathbb{R}$$ is clearly not contained in $$\mathbb{R} \setminus \{ 0 \}$$.

Can I say $$\mathbb{R} \setminus \{ 0 \} \cong \mathbb{R}^+ \times \mathbb{Z}_2$$.

• I think there should be $\Bbb{R}\backslash\{0\}$. – John Nash Sep 21 '20 at 14:56
• yes ill fix that – user486995 Sep 21 '20 at 14:57
• Yes, if you can find an isomorphism (albeit trivial) – player3236 Sep 21 '20 at 15:01

Hints: Define $$\mathbb{Z}_2$$ as the multiplicative group $$\{\pm 1\}$$ and consider a map $$\psi: \mathbb{R^+} \times \mathbb{Z}_2 \longrightarrow \mathbb{R} \backslash \{0\}$$ such that $$\psi(x, \pm 1)= \pm e^x$$. Can you check from here?

The groups $$\mathbb{R}^+$$ and $$\mathbb{Z}_2$$ are not contained in $$\mathbb{R}\setminus\{0\}$$ either.

However, you could prove that $$\mathbb{R}\setminus\{0\}\simeq ]0,+\infty[\times\{\pm1\}$$.

If you know the notion of internal direct product of subgroups, you could also prove that $$\mathbb{R}\setminus\{0\}=]0,+\infty[ \ \odot \times\{\pm1\}$$ (which implies the previous isomorphism). Since we deal with abelian groups, this amounts to check that any nonzero real number maybe written in a unique way $$\varepsilon y$$, where $$\varepsilon\in\{\pm 1\}$$ and $$y>0$$.

If you choose to prove only the isomorphism, this idea will be useful to prove it anyway.

• $\mathbb R^+$ is indeed contained in $\mathbb R \setminus \{0\}$, it is identical to what you have denoted $]0,+\infty[$. – Lee Mosher Sep 21 '20 at 15:33
• NO, it isn't. For me, $\mathbb{R}^+=[0,+\infty[$. – GreginGre Sep 21 '20 at 16:28
• While it is for you, in the context of this post is unlikely to be for the OP (and for others; the notation $\mathbb R^+$ is rather fluid). – Lee Mosher Sep 21 '20 at 16:49