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Let $G$ be a finite group, and a homomophism $f: G \rightarrow GL_n(\mathbb Z)$. Assume $f \otimes \mathbb Q: G \rightarrow GL_n(\mathbb Z) \hookrightarrow GL_n(\mathbb Q)$ is an irreducible $\mathbb Q$-representation of $G$, must $f \otimes \mathbb F_p: G \rightarrow GL_n(\mathbb Z) \rightarrow GL_n(\mathbb F_p)$ is an irreducible $\mathbb F_p$-representation of $G$ for large enough $p$?

If $f \otimes \mathbb Q$ is absolute irreducible, one can show this by character theory.

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No. For instance, let $G$ be cyclic of order $4$ and let $f:G\to GL_2(\mathbb{Z})$ send a generator to $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Then for any field $K$, $f\otimes K$ is reducible iff this matrix has an eigenvalue in $K$, i.e. iff $-1$ has a square root in $K$. So, $f\otimes\mathbb{Q}$ is irreducible, but $f\otimes\mathbb{F}_p$ is reducible for any prime $p$ that is $1$ mod $4$, and there are infinitely many such $p$.

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