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 $x \in F$?



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


