Why do we have that Hom is an exact functor in the situation described below?

Question

We are given a finite $p$-group $G$ and a finite $G$-module $M$ such that $pM=0$ (therefore $M$ is in particular a $\mathbb{F}_p$-vector space). In addition we have an arbitrary $G$-module $N$ which is also $\mathbb{F}_p$-vector space. Then we consider the short exact sequence $$0\longrightarrow M^G \overset{i}{\longrightarrow} M \overset{\pi}{\longrightarrow} M/M^G \longrightarrow 0$$ with $i$, reps. $\pi$, the natural inclusion, resp. projection.
Since $Hom(\cdot,N)$ (here $Hom(\cdot,N)=Hom_\mathbb{Z}(\cdot,N)$, that is, we consider only the group structure of the modules, i.e. $Hom(\cdot,N)$ denotes group moprhisms) is a contravariant left-exact functor, we get the exact sequence $$0\longrightarrow Hom\left(M/M^G,N\right)\longrightarrow Hom(M,N)\longrightarrow Hom\left(M^G,N\right)$$ of $G$-modules.
However there is even more, we have that $Hom(M,N)\longrightarrow Hom\left(M^G,N\right)$ is surjective. (I know it is true, but I am not able to show it.) so that the above exact sequence extends to $$0\longrightarrow Hom\left(M/M^G,N\right)\longrightarrow Hom(M,N)\longrightarrow Hom\left(M^G,N\right)\longrightarrow 0$$ I was trying to show that we have a split of $i$ (i.e. a map, say $t$, such that $t\circ i=id_{M^G}$) with something of the form $$m\in M \longmapsto \sum_{g\in G}g.m \in M^G$$ but it does not work.
Are there any better suggestions to prove this surjectivity? Thank you in advance.

Answer 1

It's not true.

For example, let $G$ be a cyclic group of order 2, $k=\mathbb{F}_2$ the field of two elements, $M=kG$ the regular $kG$-module, so that $M^G$ is the trivial $kG$-module $k$, and $N=M^G=k$ the trivial $kG$-module.