A $\mathbb Z/p\mathbb Z[G]$ submodule with no complement Let $G$ be a group acting on a set $X$ of size $n$. Suppose $G$ acts doubly transitively. If $p$ is a prime, this naturally gives a permutation representation on the  vector space over $\mathbb Z/p\mathbb Z$ with basis $\{e_{x_i}\}$, where $x_i\in X$. Elements in this vector space look like $\sum_i \alpha_i e_{x_i}$. 
We suppose $p$ divides $n$. Let $V$ be the subspace of elements such that $\sum \alpha_i =0$. Let $W$ be the submodule generated by $(1,1,1,\dots, 1)$. (This is indeed a submodule by the way we chose $p$.) I would like to show that $W$ has no $G$-stable complement in $V$.
Unfortunately, it seems one cannot mimic the proof of the converse of Maschke's theorem here, so I do not see a way forward.
 A: The following argument works provided that we make the extra assumption $p\neq2$.
Assume contrariwise that there is a $G$-equivariant homomorphism $s:V\to W$ such that $s(w)=w$ for all $w\in W$. Let us fix two elements $x,x'\in X, x\neq x'$. Because the action of $G$ is doubly transitive there exists an element $g\in G$ such that $g\cdot x= x'$ and $g\cdot x'=x$. Because $W$ is a trivial $G$-module
$$
s(e_x-e_{x'})=s(g\cdot(e_{x'}-e_x))=g\cdot s(e_{x'}-e_x)=s(e_{x'}-e_x)
=-s(e_x-e_{x'}).
$$
Hence $2s(e_x-e_{x'})=0$, so the assumption $p>2$ implies that $s(e_x-e_{x'})=0$. But vectors of the form $e_x-e_{x'}$ span all of $V$, so this implies that $s\equiv0$. This is a contradiction.

If $p=2$, then 2-transitivity gives, as above, that 
$$
s(e_x+e_{x'})=s(e_{x''}+e_{x'''})\qquad(*)
$$
for all $x,x',x'',x'''\in X$ such that $x\neq x'$ and $x''\neq x'''$. Let us fix $x_1,x_2\in X$. If $n>2$ we can find an auxiliary element $x_3\in X$. By equation $(*)$ we have
$$
\begin{aligned}
s(e_{x_1}+e_{x_2})=s(e_{x_1}+e_{x_3})+s(e_{x_2}+e_{x_3})=2s(e_{x_1}+e_{x_3})=0.
\end{aligned}
$$
This, again, implies that $s$ vanishes identically.
If $|X|=2$, then $V=W$.
