# Is $(\mathbb{R},+)$ isomorphic to a subgroup of $S_\omega$?

Is $$(\mathbb{R},+)$$ isomorphic to a subgroup of $$S_\omega$$, the group of permutations of the set of non-negative integers $$\omega$$?

• 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. – YCor Jul 15 at 12:26
• Actually the answer was already contained in the end of my previous answer to the same OP. – YCor Jul 15 at 15:10

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.)
• @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 vertex-set of its Bruhat-Tits 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. – YCor Jul 15 at 12:26
• $(\mathbb{R},+)$ 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? – Gabe Conant Jul 15 at 12:28
• 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 $2^{\kappa}$. – KConrad Jul 15 at 12:29
• 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). – Andreas Blass Jul 15 at 18:52