Taking $G$-invariants over a short exact sequence of $G$-modules Let $G$ be a finite group and
$$
 0 \to P \stackrel{\phi}{\to} M \stackrel{\psi}{\to} N \to 0
$$
be an exact sequence of $G$-modules. In Appendix B of Silverman's Arithmetic of Elliptic Curves, it is said that taking the $G$-invariants gives an exact sequence
$$
 0 \to P^G \stackrel{\phi}{\to} M^G \stackrel{\psi}{\to} N^G
$$
but $\psi: M^G \to N^G$ does not necessarily have to be surjective.
Could you give me an example for such a case? I would like to understand this better but find these definitions about $G$-modules too abstract to begin with.
 A: Hint:
Look at the sequence
$$
0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2 \to 0
$$
as $\mathbb{Z}/2$-modules. Here the action on each module is by $x \mapsto -x$ (note this makes the action on $\mathbb{Z}/2$ trivial). Notice this means the action commutes with each of the module homs as well, so this really is a sequence of $\mathbb{Z}/2$-modules.
Now take $\mathbb{Z}/2$ invariants. Do you see what you get in each case? I'll include the answer under the fold, but you should really try to compute it yourself first!

 The only element of $\mathbb{Z}$ that gets fixed is $0$, but the whole of $\mathbb{Z}/2$ is fixed. So the sequence becomes
 $$ 0 \to 0 \to 0 \to \mathbb{Z}/2 $$
 and surjectivity fails.

As an aside, one can show the $G$-invariant functor is naturally isomorphic to the functor $\text{Hom}_{\mathbb{Z}G}(\mathbb{Z}, -)$. This might give some intuition for why this functor is left exact, but not exact in general. In fact, just like we can measure the failure of exactness of $\text{Hom}$ by looking at cohomology, there is a notion of group cohomology which measures the failure of exactness for this functor.

I hope this helps ^_^
