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It can be shown that if an additive subgroup of $\Bbb R^2$ contains the unit circle, then it is necessarily all of $\Bbb R^2$. Does this also hold for a suitably smoothly deformed unit circle?

Denote by $S$ the space $\Bbb R^2\setminus\{0\}$ (the plane minus the origin) under the induced topology. Suppose $C$ is a loop that is smoothly homotopic in $S$ to the unit circle. Map $C$ to $\Bbb R^2$ using the obvious inclusion map. By abuse of notation we will also denote this loop in $\Bbb R^2$ as $C$.

If an additive subgroup of $\Bbb R^2$ contains $C$, is it necessarily all of $\Bbb R^2$?

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    $\begingroup$ The subgroup generated by any continuous image of a segment is path-connected. Any path-connected subgroup of $\mathbf{R}^n$ is a linear subspace, see ams.org/journals/proc/1969-020-01/S0002-9939-1969-0233923-X/… (see math.stackexchange.com/a/2170573/35400 for more references) $\endgroup$ – YCor Dec 15 '18 at 7:09
  • $\begingroup$ @YCor: "The subgroup generated by any continuous image of a segment is path-connected." How does one show that? In particular, I'm having trouble seeing why there is a path in the subgroup from 0 to a generator. $\endgroup$ – Nate Eldredge Dec 15 '18 at 16:25
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    $\begingroup$ @NateEldredge thanks of course you're right, this is trivially false; it holds if the generating subset is supposed to be symmetric or contains $0$. Let me fix this. Let $G$ be a subgroup of $\mathbf{R}^n$ generated by a nonempty path-connected subset $S$, and fix $s\in S$, and write $T=S+\{-s\}\subset G$. So the subgroup $V$ generated by $T$ is path-connected, hence is a subspace by the mentioned result. Then modulo $V$, $G$ is generated by $s$. So either $s\in V$ and $G=V$, or $G$ is contained in the linear image of $\mathbf{R}^k\times\mathbf{Z}\times\{0\}^{n-k-1}$ for some $k<n$. $\endgroup$ – YCor Dec 15 '18 at 16:32
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    $\begingroup$ And consequently, any path-connected subset of $\mathbf{R}^2$ that is not contained in any affine line, generates $\mathbf{R}^2$ as an additive group. More generally, any path-connected subset of $\mathbf{R}^n$ that is not contained in any affine hyperplane, generates $\mathbf{R}^n$ as an additive group. $\endgroup$ – YCor Dec 15 '18 at 16:34

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