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

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

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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.

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    $\begingroup$ $\mathbb R^+$ is indeed contained in $\mathbb R \setminus \{0\}$, it is identical to what you have denoted $]0,+\infty[$. $\endgroup$ – Lee Mosher Sep 21 '20 at 15:33
  • $\begingroup$ NO, it isn't. For me, $\mathbb{R}^+=[0,+\infty[$. $\endgroup$ – GreginGre Sep 21 '20 at 16:28
  • $\begingroup$ 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). $\endgroup$ – Lee Mosher Sep 21 '20 at 16:49

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