A more or less tautological example that shows that group cohomology is "needed" is the fact that taking invariants under finite groups in short exact sequences does not preserve exactness.
It is easy to find examples of surjections $f:M\to N$ of $G$-modules such that the induced map $M^G\to N^G$ on the invariant subspaces is not surjective.
For example, if your students are familiar with basic algebraic topology (or de Rham cohomology) you can study in detail the following situation.
If $G$ is a finite group which acts properly discontinuously on a manifold $M$, then the quotient $M/G$ is also a manifold and its de Rham cohomology $H^\bullet(M/G)$ is just the invariant subspace $H^\bullet(M)^G$ for the natural action of $G$ on $H^\bullet(M/G)$. This can be proved completely by hand.
But if the group is not finite, then this is no longer true. A minimal example is $G=\mathbb Z$ acting on $M=\mathbb R$ by translations. Then $M/G$ is a circle, which has non-trivial $1$-cohomology, so $H^1(M/G)\not\cong H^1(M)^G$.
What's wrong with this example is that $\mathbb Z$ has cohomology inpositive degrees, so we don't get isomorphisms but short exact sequences involving group cohomology. In this particular case one can be complete explicit about this.