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Let $\text{Homeo}(\mathbb{R})$ denote the group of self-homemorphisms of $\mathbb{R}$. If $G$ is a finite group, is $G$ isomorphic to a subgroup of $\text{Homeo}(\mathbb{R})$?

(Edit: this was previous denoted $\mathrm{Aut}(\mathbb{R})$, so in the comments, $\mathrm{Aut}(\mathbb{R})$ has to be interpreted as $\text{Homeo}(\mathbb{R})$.)

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    $\begingroup$ There is a normal subgroup of $\text{Aut}(\mathbb{R})$ of index $2$, namely the group $\text{Aut}^+(\mathbb{R})$ of order-preserving homeomorphisms. Thus, if $G$ is a subgroup of $\text{Aut}(\mathbb{R})$, then $G\cap \text{Aut}^+(\mathbb{R})$ is a normal subgroup of $G$ of index $1$ or $2$. For a finite subgroup of $\text{Aut}^+(\mathbb{R})$, the orbits of the induced action on $\mathbb{R}$ must just be singleton sets (just the least element in each finite orbit). Thus, the only finite groups isomorphic to a subgroup of $\text{Aut}(\mathbb{R})$ are $\{e\}$ and a cyclic group of order $2$. $\endgroup$
    – Jason Starr
    Sep 19, 2018 at 10:56
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    $\begingroup$ A quick way to see the answer is no is to note that there can be no element of order other than $4$. Indeed, if $f$ were such an element, $f^2$ would be increasing and have order $2$. But if $f(x)\neq x$, say $f(x)>x$, then $f(f(x))>f(x)>x$. $\endgroup$
    – Wojowu
    Sep 19, 2018 at 11:00
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    $\begingroup$ Thanks both, can either of zou post this as an answer so we can close this thread? $\endgroup$
    – Dominic van der Zypen
    Sep 19, 2018 at 11:04
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    $\begingroup$ It's a well known, beautiful and not terribly difficult result that a countable group embeds in $\mathrm{Aut}^+(\mathbb{R})$ (sic) if and only if it admits a left order. $\endgroup$
    – HJRW
    Sep 19, 2018 at 11:40
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    $\begingroup$ @Wojowu I'd also say, on the lines of your argument, that if $f\in\text{Aut}(\mathbb{R})$ has finite order then $f^2=\text{id}$. $\endgroup$
    – Pietro Majer
    Sep 19, 2018 at 11:56

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I am just posting my comment above as an answer.

There is a normal subgroup of $\text{Homeo}(\mathbb{R})$ of index 2, namely the group $\text{Homeo}^{+}(\mathbb{R})$ of order-preserving homeomorphisms. Thus, if $G$ is a subgroup of $\text{Homeo}(\mathbb{R})$, then $G\cap\text{Homeo}^{+}(\mathbb{R})$ is a normal subgroup of $G$ of index 1 or 2. For a finite subgroup of $\text{Homeo}^{+}(\mathbb{R})$, the orbits of the induced action on $\mathbb{R}$ must just be singleton sets (just the least element in each finite orbit). Thus, the only finite groups isomorphic to a subgroup of $\text{Homeo}(\mathbb{R})$ are $\{e\}$ and a cyclic group of order 2.

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