Control of $p$-extensions by subgroups of index coprime to $p$

Question

Let $G$ be a finite group and let $M$ be a $G$-module that is a finite abelian $p$-group. Suppose we have extensions

$1 \rightarrow M \rightarrow E_1 \rightarrow G \rightarrow 1$

and

$1 \rightarrow M \rightarrow E_2 \rightarrow G \rightarrow 1$

such that the extensions have a common restriction

$1 \rightarrow M \rightarrow F \rightarrow H \rightarrow 1$

where $F \le E_1 \cap E_2$ such that the index of $F$ in $E_1$ and $E_2$ is coprime to $p$. Then $H$ has index coprime to $p$ in $G$, so the restriction map $\mathrm{H}^2(G,M) \rightarrow \mathrm{H}^2(H,M)$ is injective, and both extensions of $G$ by $M$ give rise to the same element of $\mathrm{H}^2(H,M)$ (corresponding to the equivalence class of $F$), hence they must also give rise to the same element of $\mathrm{H}^2(G,M)$. In other words there is an equivalence of extensions $\theta:E_1 \rightarrow E_2$.

Can we choose $\theta$ so that $\theta(x) = x$ for all $x \in F$?

Answer 1

I don't think this is always possible.

Let $p=3$, $M = \langle a \rangle$ with $a^3=1$, $G = D_6 = \langle b,c \mid b^3=c^2=1, cbc=b^{-1} \rangle$, with trivial action of $G$ on $M$ and $H=\langle b \rangle$.

Let $$E_1 = \langle a,b,c \mid a^3=b^3=c^2=1, ab=ba, ac=ca, cbc=b^{-1} \rangle$$ and $$E_2 = \langle a,b,c \mid a^3=b^3=c^2=1, ab=ba, ac=ca, c(ba)c=(ba)^{-1} \rangle,$$ where $E_1$ and $E_2$ intersect in $F = \langle a,b \rangle$. So they are both split extensions of $M$ by $G$.

There is an equivalence of the extensions $E_1$ and $E_2$ with $a \mapsto a$, $b \mapsto ba$, $c \mapsto c$. But, because the automorphism of $F$ that this induces does not extend to an automorphism of $E_1$, there is no such equivalence that induces the identity on $F$.