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Let $\mathbb{F}_p$ be a finite field. There is an action (by conjugation) of $\text{GL}_2(\mathbb{F}_p)$ on the vector space of $2 \times 2$ matrices with coefficients in $\mathbb{F}_p$ that have trace zero. Is this an irreducible representation?

And is there some way to interpret this in terms of Lie theory? i.e. the representation I described is in some sense the "adjoint representation" to $\text{SL}_2(\mathbb{F}_p)$. (But obviously, this isn't a Lie group...)

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For your first question: Yes, that representation is irreducible as long as the characteristic is not 2, in which case the set of scalar matrices all have trace 0 and clearly are invariant under conjugation. To show this, observe that your representation has a basis consisting of the matrices $$\left(\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right), \left(\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right),\left(\begin{array}{cc}0 & 0 \\ 1 & 0\end{array}\right)$$ and that any invariant subspace containing one of these will also contain the others (exercise). Finally, given some arbitrary (non-zero) matrix of trace 0, you can show that any invariant subspace containing it will also contain one of those basis vectors (another exercise), and then you are done.

Now, for the second question. As you point out yourself, it is not a Lie group. It is, however, what is known as a finite group of Lie type. This means that the representation theory of this finite group is closely linked to the representation theory of the algebraic group $GL_2(K)$ where $K$ is an algebraically closed field of characteristic $p$. In fact, the finite dimensional irreducible representations of the finite group $GL_2(q)$ (now over $K$) all come from restricting certain irreducible representations of $GL_2(K)$.

To be more specific (and yet at the same time more general), let $L$ be a finite dimensional irreducible representation of $GL_n(K)$ (note the change to $n$ instead of 2). Then $L$ is the unique simple subrepresentation of a certain representation $\nabla(\lambda)$ where $\lambda = (\lambda_1,\lambda_2,\dots, \lambda_n)\in\mathbb{Z}^n$ with $\lambda_i \geq \lambda_{i+1}$. ($\nabla(\lambda)$ is obtained by starting with a 1-dimensional representation $K_{\lambda}$ for the set of lower triangular matrices where any matrix $A$ acts by the scalar $A_{1,1}^{\lambda_1}\cdots A_{n,n}^{\lambda_n}$ and inducing this to $GL_n(K)$). We will denote such a simple module by $L(\lambda)$.

If $\lambda$ further satisfies $\lambda_i - \lambda_{i+1} < q$ then restricting $L(\lambda)$ to $GL_n(q)$ gives an irreducible representation, and all finite dimensional representations of $GL_n(q)$ are obtained this way (though different $\lambda$'s can give the same representation of $GL_n(q)$.

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