Proving that a group representation is *not* a direct sum of irreducible represenations. 
Problem Statement: Let $x$ be a generator of a cyclic group $G$ of order $p$. Sending $x\mapsto \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$ defines a matrix representation $G\rightarrow GL_2(\mathbb{F}_p)$. Prove that this representation is not the direct sum of irreducible representations.

As I approach this proof I am a little confused because Maschke's Theorem states:

Every representation of a finite group $G$ is a direct sum of irreducible representations.

But since the group in question $G$ has order $p$, should that not mean that $G$ is a direct sum of irreducible representations?
I figure this must have to do with the fact that $G$ has prime order. I was trying to represent the action of $G$ as a matrix but not sure if that would be helpful, since our representation sends all $x\in\mathbb{F}_p$ to $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$
Any suggestions to help me get in the right direction?
 A: Maschke's theorem holds if the characteristic of the field doesn't divide the order of the group (you missed a very important hypothesis there!). The exercise is asking you to show that this representation is decomposable but not reducible (meaning it has a subrepresentation but it can't be written as a direct sum of irreducible representations) 
Let's look at the action of $X$ on the generic element of $\mathbb{F}_p^2$
$$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix}a \\ b\end{pmatrix} = \begin{pmatrix} a+b \\ b\end{pmatrix}$$
So we have that it can be restricted to the subspace generated by $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ (so it has a subrepresentation given by $x.a = a$), but if you consider its complement, the subspace generated by $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$, it is easy to see that the action of $x$ does not define a subrepresentation, in fact
$$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix}0\\ 1\end{pmatrix} = \begin{pmatrix} 1 \\ 1\end{pmatrix} \not\in \; \operatorname{span}\begin{pmatrix}0 \\ 1 \end{pmatrix}$$
Now, you have to show that none of the possible complements actually give you a direct sum decomposition. Take the generic vector $\begin{pmatrix}c \\ 1 \end{pmatrix}$ (all the possible complements, we scale $b$ to $1$) and see that since
$$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix}c\\ 1\end{pmatrix} = \begin{pmatrix} c+1 \\ 1 \end{pmatrix} $$
those two vectors always generate all the space, and so there is no possible direct sum decomposition of $\mathbb{F}_p^2$ into subrepresentations.
