Is $(\mathbb{R},+)$ isomorphic to a subgroup of $S_\omega$, the group of permutations of the set of nonnegative integers $\omega$?
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$\begingroup$ I think this is a duplicate, of some post here or MathSE. BtW you mean isomorphic as abstract groups, since it's clearly impossible with the usual topologies. $\endgroup$ – YCor Jul 15 at 12:26

1$\begingroup$ Actually the answer was already contained in the end of my previous answer to the same OP. $\endgroup$ – YCor Jul 15 at 15:10
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See Theorem 4.3 of this paper by De Bruijn. Any abelian group of order $2^\kappa$ can be embedded in $Sym(\kappa)$ when $\kappa$ is infinite. (There is also an addendum to the paper which corrects some error in the proof.)


11$\begingroup$ @KConrad quite immediately, just using $\kappa=\omega$. Actually this case is obvious from many standard facts. For instance, $G=PSL_2(\mathbf{Q}_p)$ acts faithfully on the vertexset of its BruhatTits tree, a countable set. Then $G$ contains subgroups isomorphic to $\mathbf{Q}_p$, which is isomorphic to $\mathbf{R}$ as abstract group. Actually this can be refined to show that every linear group over a field of cardinal $\le c$ acts faithfully on a countable set. Alternatively, $\mathbf{R}\simeq\mathbf{Q}^\mathbf{N}$ acts faithfully on $\mathbf{Q}\times\mathbf{N}$ in the obvious way. $\endgroup$ – YCor Jul 15 at 12:26

2$\begingroup$ $(\mathbb{R},+)$ is an abelian group of order $2^{\aleph_0}$ and so embeds in $S_\omega$ by the theorem I quoted. Unless I’m misreading something? $\endgroup$ – Gabe Conant Jul 15 at 12:28

1$\begingroup$ Oops, I misread $\kappa$ as $k$ and thought the answer was about abelian subgroups of order $2^k$ for each nonnegative integer $k$ instead of one subgroup of order $2^{\kappa}$. $\endgroup$ – KConrad Jul 15 at 12:29

7$\begingroup$ In the cited paper of de Bruijn, he mentions (p. 561) that the special case of the reals and Sym$(\omega)$ was proved by Karrass and Solitar a year earlier (1956). $\endgroup$ – Andreas Blass Jul 15 at 18:52