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Let $\langle a, b \rangle = F_2$ be a two-generator free group and $\hat{F_2}$ be its profinite completion. Is there an element $c\in \hat{F_2}$ such that $\langle a, b, c\rangle \le \hat{F_2}$ is isomorphic to the 3-generator abstract free group $F_3$?

I posted the same question on Math StackExchange(link), but my question hasn't been answered yet. I think the construction or the proof of nonexistence would be simple, but I couldn't make it yet.. apologies if I violated some etiquette.

Edited: Assuming there is such an element $c$, I tried to make a continuous epimorphism $f:\hat{F_2} \to \hat{H}$, where $H$ is a 3-generator free group with $H=\langle h_1, h_2, h_3\rangle$ and $f(a)=h_1, f(b)=h_2, f(c)=h_3$. Then, since we have a projection from $\hat{F_3}$ to $\hat{F_2}$, we attain a contradiction. However, I cannot convince myself $f$ is well constructed.

It is impossible; see comments by William Chen.

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    $\begingroup$ I suspect this is possible - If you consider the analogous question for free abelian groups (say pro-$p$ for simplicity), you can consider the elements $a = (1,0)$, $b = (0,1)$, and $c = (t,0)$ in $\mathbb{Z}_p\times\mathbb{Z}_p$ where $t\in\mathbb{Z}_p$ is transcendental over $\mathbb{Z}$. Then $a,b,c$ are $\mathbb{Z}$-linearly independent so generate an abstract free abelian group of rank 3. I suspect a similar example can be constructed in your case with a little more work. $\endgroup$ – Will Chen Jun 16 '18 at 7:53
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    $\begingroup$ In your example with $f : \widehat{F}_2\rightarrow\widehat{H}$, $f$ cannot be continuous since if it were then $h_3$ would be in the closure of the group generated by $h_1,h_2$, which it clearly isn't. $\endgroup$ – Will Chen Jun 16 '18 at 7:57
  • $\begingroup$ Here's a partial recipe to get such a group. Find a profinite group $K$ with three elements $a',b',c'$ such that $\langle a',b'\rangle$ is dense and $(a',b',c')$ is free. If so, we can "pull them back" to $\hat{F_2}$. Namely considering the unique homomorphism $\hat{F_2}\to K$ mapping $a\mapsto a'$, $b\mapsto b'$; density implies that it's surjective and hence we can choose $c$ as any preimage of $c'$. The point is that it's enough to choose $K$ among many better-understood profinite groups. $\endgroup$ – YCor Jun 16 '18 at 8:08
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Yes. Here's a recipe to get such a group. Find a profinite group $K$ with three elements $a',b',c'$ such that $\langle a',b'\rangle$ is dense and $(a',b',c')$ is a free family. If so, we can "pull them back" to $\hat{F_2}$. Namely considering the unique homomorphism $\hat{F_2}\to K$ mapping $a\mapsto a'$, $b\mapsto b'$; density implies that it's surjective and hence we can choose $c$ as any preimage of $c$.

Let's now find $K$ (there are many alternative ways). Consider the three matrices $$A=\begin{pmatrix}4 & 5\\3 & 4\end{pmatrix},B=\begin{pmatrix}7 & 8\\6 & 7\end{pmatrix},C=\begin{pmatrix}4 & 9\\3 & 7\end{pmatrix}.$$ We'll successively view them as real, and as 3-adic matrices.

These are loxodromic elements in $\mathrm{SL}_2(\mathbf{R})$, with pairwise disjoint pairs of fixed points in the projective line. Hence they have the property that for any large enough $n$, say $n\ge N_0$, the powers $A^n,B^n,C^n$ freely generate a free group.

This was the use of real numbers. Now view them as 3-adic matrices. Observe that all of them belong to the subgroup $K$ of $\mathrm{SL}_2(\mathbf{Z}_3)$ of those matrices $\begin{pmatrix}x & y\\z & t\end{pmatrix}$ such that $x-1,t-1,z$ are in $3\mathbf{Z}_3$ (i.e. the inverse image of the upper unipotent subgroup in reduction modulo 3).

The profinite group $K$ is actually a pro-$3$-group. Its elementary abelianization is given by the homomorphism $f:\begin{pmatrix}1+3a & b\\3c & 1+3d\end{pmatrix}\mapsto (b,c)\in (\mathbf{Z}/3\mathbf{Z})^2=:L$. A basic fact in pro-$p$-groups is that a subgroup $\Gamma$ of $K$ is dense iff $f(\Gamma)=L$. Since $f(A)=(1,2)$, $f(B)=(2,2)$ and $f(C)=(1,0)$ which are pairwise non-collinear, this applies to the subgroup generated by any of the three pairs in $\{A,B,C\}$.

Moreover, for any $n$ with $n$ coprime to 3 and any $u\in\{A,B,C\}$, the cyclic subgroups generated by $u$ and $u^n$ have the same closure in $K$. If furthermore $n\ge N_0$, we deduce that $\langle A^n,B^n\rangle$ is still dense and $(A^n,B^n,C^n)$ freely generates a free subgroup. So this is the desired pair $(a',b',c')$.

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