= Solution
A complex representation of $G$ is a complex <vector space> $V$ together with a homomorphism
$$
\rho:G\longrightarrow\operatorname{GL}(V).
$$
It is faithful when $\ker\rho=\{e\}$, equivalently when $\rho$ is injective. Every finite <group> has a faithful complex representation: in the <regular representation>, $G$ permutes the <basis> $\{e_h:h\in G\}$ by
$$
\rho(g)e_h=e_{gh},
$$
and a <group> element fixing every <basis> <vector> must be the identity.
If $g$ is conjugate to $g^k$, then $\rho(g)$ is similar to $\rho(g^k)=\rho(g)^k$. Thus the two <matrices> have the same <spectrum>, while an <eigenvalue> $\lambda$ of $\rho(g)$ gives the <eigenvalue> $\lambda^k$ of $\rho(g)^k$. Hence
$$
\lambda\in S(g)\quad\Longrightarrow\quad\lambda^k\in S(g).
$$
Now take a $p$-cycle $g\in S_p$. It is conjugate to $g^k$ for every $1\leq k<p$. Faithfulness makes $\rho(g)$ have order $p$, so its <spectrum> contains a nontrivial $p$th root of unity $\lambda$. The preceding implication puts all the $p-1$ distinct values
$$
\lambda,\lambda^2,\ldots,\lambda^{p-1}
$$
in the <spectrum>. Therefore $\dim V\geq p-1$. This is the <spectrum orbit bound for a faithful symmetric-group representation>.
Finally let $G$ be the <group> of all permutations of $\mathbb N$. It has no faithful finite-dimensional complex representation. For every $n$, $G$ contains a <subgroup> $(C_2)^n$ generated by $n$ disjoint transpositions. In a $d$-dimensional complex representation, commuting involutions are simultaneously diagonalizable and hence map into the <group> of diagonal sign <matrices>, which has order $2^d$. A faithful restriction would require $2^n\leq2^d$ for every $n$, an impossibility.
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