Is any continuous Galois-representation a direct summand of a permutative one? Lets $k$ be a (perfect) field and $\Gamma=\text{Gal}(k^{s}/k)$ its absolute Galois group.
Let $F$ be a field of characteristic zero (e.g. $F=\mathbb{Q}$), condsider
now the category $\textrm{Rep}(\Gamma)$of continuous representations of $\Gamma$ in finite dimensional $F$-vectorspaces. I now call such a represensation $\textit{permutative}$ if the corresponding vectorspace $V$ admits a basis, such that $\Gamma$ acts on $V$ via permutations of the basis.
Hence the subcategory of $\textrm{Rep}(\Gamma)$ consisting of permutative representations is equivalent to the category of $\Gamma$-sets.
My question is now:
If $\rho_{1}\in\text{Rep}(\Gamma)$, is there another $\rho_{2}\in\text{Rep}(\Gamma)$ such that $\rho_{1}\oplus\rho_{2}$ is a permutative representation?
 A: First, let me remark that the category of permutation representations in characteristic 0 is very far from being equivalent to the category of $\Gamma$-sets, since non-isomorphic $\Gamma$-sets can induce isomorphic representations. This is already true on the finite level: you might want to google for "Brauer relations".
The answer to your question is 'yes'. A continuous Galois representation factors through a finite quotient and any representation of a finite group in characteristic zero is projective. E.g. if it is irreducible, then it is always a direct summand of the regular representation.
A: Any continuous representation $V$ of $\Gamma$ will factor through some finite quotient of $\Gamma$, say $\Gamma/N=G$.  We can then choose an epimorphism $F[G]^N\to V$ for some $N$, and let $V'$ be the kernel.  By the usual averaging argument this will split to give $V\oplus V'\simeq F[G]^N$, and clearly $F[G]^N$ is permutative.
Incidentally, it is not actually true that the permutative subcategory of $Rep(\Gamma)$ is equivalent to that of continuous $\Gamma$-sets.  The obvious reason is that the latter is not an additive category.  More subtly, if $G$ is elementary abelian of order $4$, and $A$, $B$ and $C$ are the three subgroups of order $2$, then the $G$-sets $G/A\amalg G/B\amalg G/C$ and $G/1\amalg G/G\amalg G/G$ are non-isomorphic but give isomorphic representations.
