TITLE: Counterexample for Maschke's lemma
QUESTION [5 upvotes]: I'm trying to relax conditions and come up with a counterexample to Maschke's lemma in such a case. For example, with $k = \mathbb Z/p\mathbb Z$, I'm considering the two dimensional representation of the cyclic group $G = \mathbb Z/p\mathbb Z$ over $k$ deﬁned as $a$ going to the matrix $
   M=
  \left[ {\begin{array}{cc}
   1 & a \\
   0 & 1 \\
  \end{array} } \right]
$ 
where $a$ is a generator of $G$. What will be a suitable subspace to show that Maschke’s theorem does not hold?

REPLY [5 votes]: Try span of $[1\;0]^t$, i.e.
$$ V =  \left\{ \begin{bmatrix}a\\0\end{bmatrix} \mid a\in\mathbb Z/p\mathbb Z \right\}. $$
You can simply compute everything, but in fact there is a direct way to see that there can not be a complementary representation and that hence Maschke's theorem must fail. If $V$ had a complement, then the matrix $M$ would be diagonalizable, but you may remember from a class in linear algebra that it isn't, because the algebraic multiplicity of the eigenvalue $1$ (which is $2$) is larger than the geometric multiplicity (which is $1$).