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Recall that an abstract a group $G$ is divisible if for every positive integer $n$, the map $x \mapsto x^n$ is surjective on $G$.

It is easy to show that any quotient of a divisible commutative group is again divisible. Moreover, no nontrivial finite group can be divisible, since if $|G| = n$ then $x \mapsto x^n$ sends every element to the identity.

It follows that a divisible commutative group has no proper finite index subgroups, so it has trivial profinite completion.

Note that the group $(\mathbb{R},+)$ is divisible, hence so is its quotient $S^1 = (\mathbb{R},+)/(\mathbb{Z},+)$. Therefore the profinite completion of $S^1$ is the trivial group.

Added: Having said this much, I might as well add a little more to show that even a divisible non-commutative group has trivial profinite completion. To see this, note first that as above a divisible group can have no proper finite index normal subgroups, and second that if a group $G$ has a subgroup $H$ with $1 < [G:H] < \infty$, then it also has a normal subgroup $H'$ with $1 < [G:H'] < \infty$: indeed by orbit-stabilizer considerations $H$ has only finitely many distinct conjugates $g H g^{-1}$ in $G$, and we may take $H'$ to be the intersection of all these conjugates.

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Recall that an abstract group $G$ is divisible if for every positive integer $n$, the map $x \mapsto x^n$ is surjective on $G$.

It is easy to show that any quotient of a divisible commutative group is again divisible. Moreover, no nontrivial finite group can be divisible, since if $|G| = n$ then $x \mapsto x^n$ sends every element to the identity.

It follows that a divisible commutative group has no proper finite index subgroups, so it has trivial profinite completion.

Note that the group $(\mathbb{R},+)$ is divisible, hence so is its quotient $S^1 = (\mathbb{R},+)/(\mathbb{Z},+)$. Therefore the profinite completion of $S^1$ is the trivial group.