$\def\Gal{\mathrm{Gal}}$
$\def\Res{\mathrm{Res}}$
$\def\GL{\mathrm{GL}}$
$\def\F{\mathbf{F}}$
$\def\Q{\mathbf{Q}}$

**Edited to include more details.**

Let $K$ be a field whose characteristic is prime to the order of $G$.
The algebra $K[G]$ is a product of matrix algebras over division rings.
Given any *absolutely irreducible* character $\chi$ with coefficients in $L/K$, the question of whether $\chi$ may be realized over $L$ is equivalent to whether the corresponding division algebra $D/K$ splits over $L$.

Suppose that $K = k$ is a finite field. Then, by Weddeburn's theorem, there are no non-trivial division algebras and irreducible characters with values in $k$ have models over $k$.

Suppose that $K$ is a number field. For all but finitely many places $v$ of $K$, $D \otimes_K K_v$ is a matrix algebra. So, at least away from finitely many places, irreducible characters with values in $K$ have models over $K_v$.

**Example:** Let $G = Q_8$ be the quaternion group, and let $\chi$ denote the absolutely irreducible faithful character of degree $2$. Then $\chi$ is valued in $\Q$. However, the corresponding quaternion algebra $D$ is ramified at $2$ and $\infty$. Hence $G$ does not have a faithful representation over either $\mathbf{R}$ nor $\mathbf{Q}_2$, but it does for $\Q_p$ and all odd primes $p$, because $D \otimes \Q_p = M_2(\Q_p)$ in those cases.

In the example above, we see that the obstruction arises for finite places only at the prime $2$ which divides $|G|$. Let us show that there is no obstruction for any $v$ such that the residue characteristic is finite and prime to $|G|$. In such cases, there is a bijection between absolutely irreducible characters in characteristics zero and $p$ given by reduction modulo $p$. Let $\chi$ be an irreducible character with values in a local field $E$. It preserves a lattice $\mathcal{O}$, and gives rise to an irreducible character over the finite field $k$, which has a model over $k$ by the remarks above. Hence (because the characters are in bijection) it suffices to prove the following: any representation:
$$G \rightarrow \GL_n(k)$$
with image $H$ admits a lift to $\GL_n(W(k))$, since $W(k)[1/p]$ will be necessarily be a subfield of $E$. (Not surprisingly, we see that all representations have models over unramified extensions when $p \nmid |G|$, since the characters are all valued in $\Q(\zeta_{m})$ where $m$ is the exponent of $G$.)
There is a projection $\GL_n(W(k)) \rightarrow \GL_n(k)$; let $\Gamma$ denote the inverse image of $H$. Since $H$ has order prime to $p$ and the kernel of $\Gamma \rightarrow H$ is pro-$p$, by Schur-Zassenhaus there is a splitting $H \rightarrow \Gamma$ which gives the required lift.

Suppose now that $\eta$ is a general (genuine) character of $G$ over a local field $K/\Q_p$ with values generate the field $L/K$, and suppose that
$p$ does not divide $|G|$. We prove that $\eta$ has a model over $L$. As noted above, $L/K$ is unramified, so in particular is Galois and $\Gal(L/K)$ is cyclic.
The result is true for
irreducible characters from the discussion above.
Suppose that $\chi$ is an irreducible constituent of $\eta$. Since $\Gal(L/K)$ fixes $\eta$,
it follows that the $[L:K]$ distinct characters $\sigma \chi$ occur inside $\eta$.
Hence

$$\eta = \phi + \sum_{\Gal(L/K)} \sigma \chi,$$
where $\eta$ is a genuine character of lower dimension. Hence it suffices to note that
$$\bigoplus \sigma \chi$$
is defined over $K$, because if $V/L$ is a realization of $\chi$, then the above is
$\mathrm{Res}_{L/K}(V)$, where $L/K$ is thought of as a $[L:K]$-dimensional vector
space in the usual way. The result follows by induction.

**Example:** Suppose that $p \equiv -1 \mod 4$, and let $\chi$ be a faithful character of $\mathbf{Z}/4\mathbf{Z}$. Then $K = \Q_p(\chi)$ is unramified over $\Q_p$ of degree $2$, and the representation $\chi + \sigma \chi$ is sends a generator to $i \in K$ thought of as a vector space over $\Q_p$. If one chooses the basis $\{1,i\}$ of $K/\Q_p$, this is just the matrix:
$$\left( \begin{matrix} 0 & 1 \\\ -1 & 0 \end{matrix} \right).$$

**Conclusion:** Hence, at least for primes $p$ not dividing $|G|$, there is an injection from $G$ to $\GL_n(\Q_p)$ if and only if there is a faithful character $\eta$ with
values in a field $K$ which has a prime of norm $p$. Since $K$ will be abelian and ramified only at primes dividing $|G|$, this is equivalent to asking that $K$ split completly at $p$. So the set $S(G)$ (up to primes dividing $|G|$) is the union of primes which split completely in some finite number of fields determined by the faithful characters of degree $n$. If one restricts to a fixed character $\eta$, then $S_{\eta}(G)$ is indeed Galoisian.

Providing that $\eta$ has at least one faithful character of degree $n$, then $S(G)$ has rational positive density, answering 1. As for 2, it is not *strictly* Galoisian according to the definition of the previous question, since that required that the set be the set of primes
which split completely in a *single* field. For example, one can take
$$G = \mathbf{Z}/12 \mathbf{Z}$$
and $n = 2$. Then $G$ is a subgroup of $\mathbf{GL}_2(\Q_p)$ for $p > 3$ if and only if $p \equiv 1,5,7 \mod 12$, but not $11 \mod 12$, and this is not the set of primes which splits completely in any field $L$. In this case, if $\chi$ is a faithful character of $G$ of degree $1$, then $\chi + \chi^5$ and $\chi + \chi^{-1}$ are faithful characters with values in $\Q(\sqrt{-1})$ and $\Q(\sqrt{-3})$ respectively.

If $G$ has a faithful character of degree $n$ with values in $\Q$, then we see that $G$ embeds into $\GL_n(\Q_p)$ for all but finitely many primes $p$. The converse is quite possibly false, however; one could imagine $G$ having faithful characters of degree $n$ with values in $\Q(\sqrt{2})$, $\Q(\sqrt{3})$, and $\Q(\sqrt{6})$ but not in $\Q$ (although I don't have an example off the top of my head.)

Local Fieldsor to: J.W.S. Cassels, An embedding theorem for fields. Bull. Austral. Math. Soc. 14 (1976), 193-198. – Pete L. Clark Apr 23 '13 at 3:25