I'm working in a problem and in the order to solve it, I'm stuck on a claim.
Suppose that each $2$-generated closed subgroup of a profinite group $G$ is prosoluble. Is it true that $G$ is prosoluble?
I cannot find a counterexample.
Given a $2$-generated closed subgroup of $G$, we can write $$\overline{\langle x,y \rangle} = \varprojlim \pi_i(\overline{\langle x,y \rangle}) = \varprojlim \overline{\pi_i(\langle x,y \rangle)}$$ where $\pi_i: G \to G_i$ are the projections. Writting $$G = \varprojlim G_i$$ we have that $\overline{\pi_i(\langle x,y \rangle)}$ is a soluble subgroup of $G_i$. My knowledge in soluble groups is not very large, so... is there a way to use it in order to show that $G_i$ is soluble?
UPDATE. I found a result from Thompson saying that a finite group $G$ is soluble if and only if every $2$-generated subgroup is soluble.
So, if $\langle x_i,y_i \rangle$ is a $2$-generated subgroup of $G_i$, then there is a correspondent $\overline{\langle x,y \rangle}$ $2$-generated closed subgroup of $G$ such that $\pi_i(\overline{\langle x,y \rangle}) = \langle x_i,y_i \rangle$. So, $\langle x_i, y_i \rangle$ is soluble, which implies $G_i$ soluble.
Given a $2$-generated closed subgroup of $G$, we can write $$\overline{\langle x,y \rangle} = \varprojlim \pi_i(\overline{\langle x,y \rangle}) = \varprojlim \overline{\pi_i(\langle x,y \rangle)}$$ where $\pi_i: G \to G_i$ are the projections. Writting $$G = \varprojlim G_i$$ we have that $\overline{\pi_i(\langle x,y \rangle)}$ is a soluble subgroup of $G_i$.
There is a result from Thompson saying that a finite group $G$ is soluble if and only if every $2$-generated subgroup is soluble.
So, if $\langle x_i,y_i \rangle$ is a $2$-generated subgroup of $G_i$, then there is a correspondent $\overline{\langle x,y \rangle}$ $2$-generated closed subgroup of $G$ such that $\pi_i(\overline{\langle x,y \rangle}) = \langle x_i,y_i \rangle$. So, $\langle x_i, y_i \rangle$ is soluble, which implies $G_i$ soluble.