Let G be a group, p a prime number , $\mathbb{F}\_p$ finite field. Suppose we have a representation $\rho\_6$ of G into $GL\_n(\mathbb{F}\_{p^6})$ , a representation $\rho\_2$ of G into $GL\_n(\mathbb{F}\_{p^2})$, a representation $\rho\_3$ of G into $GL\_n(\mathbb{F}\_{p^3})$, Do we have the following statement ? : If $\rho\_6=\rho\_2\otimes \mathbb{F}\_6$, $\rho\_6=\rho\_3\otimes \mathbb{F}\_6$ , then there exist a representaion $\rho\_1$ of G into $GL\_n(\mathbb{F}\_p)$, such that $\rho\_6=\rho\_1\otimes\mathbb{F}\_{p^6}$.


is this statment for any coprime number m,n replace of 2,3 , and mn replace of 6 true? Thank you!



representation over finite field and field extension ? 
The answer is yes if $\rho\_6$ is absolutely irreducible. 
Here is a proof when $p>n$ using pseudo-characters. The trace of $\rho\_j$,
for $j=2,3,6$, is a pseudo-character $T\_j$ of $G$ to $\mathbb F\_{p^j}$ of dimension $n$.
For $g$ in $G$, we have $T\_6(g)=T\_2(g)=T\_3(g) \in \mathbb F\_{p^2} \cap \mathbb F\_{p^3} = \mathbb F\_p$. Hence $T\_6$ is a pseudo-character absolutely irreducible of $G$ to 
$\mathbb F\_p$ and such a pseudo-character over a finite field always comse from a representation $\rho\_1$, which clearly satisfies your condition since two absolutely irreducible representations with the same trace are isomorphic over *any* field.


Of course the same argument work when 2,3 are replaced by any relatively prime integers.
Also I have supposed $p>n$ just for the convenience of using the classical theory of pseudo-characters, which works well under this condition, but in the general case Chenevier has a general theory (that he calls "determinant" -- see on his webpage at polytechnique)
which works without this condition, and with which the proof above can be adapted without difficulties.


On the other hand, the hypothesis absolutely irreducible is necessary for the method. I
don't know if it is necessary for your question...


