Why is every l-adic Galois representation
$$G\_{\mathbb{Q}\_p}\rightarrow GL\_n(\mathbb{Q}\_{l})$$
conjugate to one over the l-adic integers?
$$G\_{\mathbb{Q}\_p}\rightarrow GL\_n(\mathbb{Z}\_{l})$$



Why is every l-adic Galois representation conjugate to one over the l-adic integers?
Recall that all ${\mathbb Z}\_l$-lattices in the ${\mathbb Q}\_l$-vector space ${\mathbb Q}\_l^n$ are conjugate in ${\rm GL}\_n ({\mathbb Q}\_l )$. Since the ${\rm GL}\_n ({\mathbb Q}\_l )$ stabilizer of the standard lattice ${\mathbb Z}\_l^n$ is ${\rm GL}\_n ({\mathbb Z}\_l )$, all lattices in ${\mathbb Q}\_l^n$ have open stabilizer in ${\rm GL}\_n ({\mathbb Q}\_l )$. Moreover a finite sum of ${\mathbb Z}\_l$-lattices is again an ${\mathbb Z}\_l$-lattice. 


Now take $\rho$ : $G\_{{\mathbb Q}\_p}\longrightarrow {\rm GL}\_n ({\mathbb Q}\_l )$ a continuous Galois representation. 
Let $L$ be any ${\mathbb Z}\_l$-lattice in ${\mathbb Q}\_l^n$. Then its stabilizer $H$ in $G\_{{\mathbb Q}\_p}$ is open and the coset set $G\_{{\mathbb Q}\_p}/H$ is finite, since $G\_{{\mathbb Q}\_p}$ is profinite. Now


$$
M := \sum\_{g\in G\_{{\mathbb Q}\_p}/H} g.L
$$


is a Galois stable lattice in ${\mathbb Q}\_l^n$. We have proved that a conjugate of $\rho (G\_{{\mathbb Q}\_p})$ lies in ${\rm GL}\_n ({\mathbb Z}\_l )$. 


