It is true that if $M$ is any (solvable) group with $E \lhd M$ and $M/E \cong G$, (with the action of $G$ on $E$ specified by the given irreducible module action), then $E$ is complemented in $M$, and all complements to $E$ are conjugate. In fact, this does not require solvability of $G,$ only $p$-solvability. All this is well-known, but I outline the proof: Note that $O_{p}(G) = 1$ since $G$ acts irreducibly on $E$, so that $O_{p}(M) = E.$ Let $K = O_{p,p^{\prime}}(M)$. Then $K$ has a unique conjugacy classes of Hall $p^{\prime}$-subgroups, say one of these is $L$, so by a Frattini-like argument we have $M = KN_{M}(L)$.

Hence $M = ELN_{M}(L) = EN_{M}(L).$ Now $E$ is a minimal normal subgroup of $M$ since $G$ acts irreducibly on $E.$ However $E \cap N_{M}(L) \lhd M$. We can't have $E \leq N_{M}(L)$, otherwise $[E,L] = 1$, whereas $G$ acts faithfully on $E$. Hence $E \cap N_{M}(L) = 1$ and $G \cong N_{M}(L)$. This argument actually shows that every complement to $E$ in $M$ has the form $N_{M}(L)$ for some Hall $p^{\prime}$-subgroup $L$, of $O_{p,p^{\prime}}(M)$ and $L$ is unique up to $M$-conjugacy, so all complements to $E$ are $M$-conjugate.

It is true that if $M$ is any (solvable) group with $E \lhd M$ and $M/E \cong G$, (with the action of $G$ on $E$ specified by the given irreducible module action), then $E$ is complemented in $M$, and all complements to $E$ are conjugate. In fact, this does not require solvability of $G,$ only $p$-solvability. All this is well-known, but I outline the proof: Note that $O_{p}(G) = 1$ since $G$ acts irreducibly on $E$, so that $O_{p}(M) = E.$ Let $K = O_{p,p^{\prime}}(M)$. Then $K$ has a unique conjugacy classes of Hall $p^{\prime}$-subgroups, say one of these is $L$, so by a Frattini-like argument we have $M = KN_{M}(L)$.

Hence $M = ELN_{M}(L) = EN_{M}(L).$ Now $E$ is a minimal normal subgroup of $M$ since $G$ acts irreducibly on $E.$ However $E \cap N_{M}(L) \lhd M$. We can't have $E \leq N_{M}(L)$, otherwise $[E,L] = 1,$ whereas $G$ acts faithfully on $E$. Hence $E \cap N_{M}(L) = 1$ and $G \cong N_{M}(L)$. This argument actually shows that every complement to $E$ in $M$ has the form $N_{M}(L)$ for some Hall $p^{\prime}$-subgroup $L$, of $O_{p,p^{\prime}}(M)$ and $L$ is unique up to $M$-conjugacy, so all complements to $E$ are $M$-conjugate.

It is true that if $M$ is any (solvable) group with $E \lhd M$ and $M/E \cong G$, (with the action of $G$ on $E$ specified by the given irreducible module action), then $E$ is complemented in $M$, and all complements to $E$ are conjugate. In fact, this does not require solvability of $G,$ only $p$-solvability. All this is well-known, but I outline the proof: Note that $O_{p}(G) = 1$ since $G$ acts irreducibly on $E$, so that $O_{p}(M) = E.$ Let $K = O_{p,p^{\prime}}(M)$. Then $K$ has a unique conjugacy classes of Hall $p^{\prime}$-subgroups, say one of these is $L$, so by a Frattini-like argument we have $M = KN_{M}(L)$. Hence $M = ELN_{M}(L) = EN_{M}(L).$ Now $E$ is a minimal normal subgroup of $M$ since $G$ acts irreducibly on $E$, However $E \cap N_{M}(L) \lhd M$. We cant' have $E \leq N_{M}(L)$, otherwise $[E,L] = 1$, whereas $G$ acts faithfully on $E$. Hence $E \cap N_{M}(L) = 1$ and $G \cong N_{M}(L)$. This argument actually shows that every complement to $E$ in $M$ has the form $N_{M}(L)$ for some Hall $p^{\prime}$-subgroup $L$, of $O_{p,p^{\prime}}(M)$ and $L$ is unique up to $M$-conjugacy, so all complements to $E$ are $M$-conjugate.

It is true that if $M$ is any (solvable) group with $E \lhd M$ and $M/E \cong G$, (with the action of $G$ on $E$ specified by the given irreducible module action), then $E$ is complemented in $M$, and all complements to $E$ are conjugate. In fact, this does not require solvability of $G,$ only $p$-solvability. All this is well-known, but I outline the proof: Note that $O_{p}(G) = 1$ since $G$ acts irreducibly on $E$, so that $O_{p}(M) = E.$ Let $K = O_{p,p^{\prime}}(M)$. Then $K$ has a unique conjugacy classes of Hall $p^{\prime}$-subgroups, say one of these is $L$, so by a Frattini-like argument we have $M = KN_{M}(L)$.

Hence $M = ELN_{M}(L) = EN_{M}(L).$ Now $E$ is a minimal normal subgroup of $M$ since $G$ acts irreducibly on $E.$ However $E \cap N_{M}(L) \lhd M$. We can't have $E \leq N_{M}(L)$, otherwise $[E,L] = 1$, whereas $G$ acts faithfully on $E$. Hence $E \cap N_{M}(L) = 1$ and $G \cong N_{M}(L)$. This argument actually shows that every complement to $E$ in $M$ has the form $N_{M}(L)$ for some Hall $p^{\prime}$-subgroup $L$, of $O_{p,p^{\prime}}(M)$ and $L$ is unique up to $M$-conjugacy, so all complements to $E$ are $M$-conjugate.