Let $p$ be an odd prime number, $G$ a finitely generated nonabelian profinite group, $L \lhd_o G$ a pro-$p$ group with $[G : L] = 2$. Suppose that there is a continuous surjection from $G$ onto a free pro-$p$ group of rank $d(G)$. Must $L$ be free pro-$p$ ?
Here, $d(G)$ is the smallest cardinality of a generating set of $G$.
No.
Let $G = D_p \times F$, where $D_p = \langle a,b \mid\ a^p = 1,\ b^2 = 1,\ a^b \ (:=b^{-1}ab)=a^{-1}\rangle$ is the dihedral group of order $2p$ and $F$ is the free pro-$p$ group on two generators, say, $x,y$. Then $L = \langle a \rangle \times F$ is of index $2$ in $G$, pro-$p$, but not free, and there is a surjection $G \to F$. Since $H := \langle ax, by \rangle$ is clearly of rank $2$, it suffices to show that $H = G$, i.e., that $a,b,x,y \in H$.
As $b,y$ are of coprime (supernatural) orders and commute, $\langle b y \rangle = \langle b \rangle \times \langle y \rangle$, so $b, y \in H$. Thus $x a^{-1} = a^{-1} x = (ax)^b \in H$. Therefore $x^2 = (x a^{-1}) (a x) \in H$ and $a^2 = (a x) (a^{-1} x)^{-1} \in H$. As $2$ is prime to the orders of $x$ and $a$, we have $\langle x^2 \rangle = \langle x \rangle$ and $\langle a^2 \rangle = \langle a \rangle$, so $x, a \in H$.
