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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?

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    $\begingroup$ 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 probably 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) $\endgroup$ – AnalysisStudent0414 Feb 7 '16 at 19:05
  • $\begingroup$ Oh, I see. We haven't learned about characteristics yet, and my professor never mentioned that part of the theorem, which is why I was confused. But now it makes sense, because the characteristic would be $p$. I am confused how I would find a sub-representation of $G$, would I need to define a linear map that always sends $x$ to $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$? Or am I treating $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ as the group acting on elements of $\mathbb{F}_p$? I'm sorry representation theory is still kind of over my head (...still trying to make sense of it all) $\endgroup$ – yung_Pabs Feb 7 '16 at 19:14
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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.

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  • $\begingroup$ Ahh, this makes much more sense. So $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ has order $p$ and represents the action of $G$ on $\mathbb{F}_p$. Let me make sure I understand. So because there are no two $1$-dimensional sub-representations which are $G$-invariant (only the subspace generated by $e_1$ of dimension $1$ and the subspace generated by $\begin{bmatrix} c & 1 \end{bmatrix}^T$ of dimension $2$) then we cannot find a direct sum of irreducible representations? $\endgroup$ – yung_Pabs Feb 7 '16 at 19:35
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    $\begingroup$ Yes, note that the subspace generated by $(c\; \; 1)^T$ of dimension $2$ is, well, the whole space $\endgroup$ – AnalysisStudent0414 Feb 7 '16 at 19:40

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