= Solution
We prove the <primitive three-cycle criterion> through supports. First, suppose a <subgroup> contains $A_U$, the <alternating group> on a set $U$ of at least three letters, and a <three-cycle> with support $E$ meeting $U$.
If $E$ adds one letter, it has two letters in $U$. The group $A_U$ is transitive on unordered pairs of $U$: for three letters its cyclic action permutes the three pairs, and for at least four letters one can adjust the parity of a <permutation> sending one pair to another while preserving its target pair. Conjugating the new cycle, and using inverses, supplies every <three-cycle> on two old letters and the new letter. Together with $A_U$, these generate $A_{U\cup E}$.
If $E$ adds two letters, write its cycle as $\tau=(a\,b\,c)$ with $a\in U$. Choose $a'\in U\setminus\{a\}$ and conjugate within $A_U$ to obtain $\tau'=(a'\,b\,c)$. Then
$$
\tau(\tau')^{-1}=(a\,b\,a').
$$
The preceding case first adds $b$; applying it again to $\tau$ adds $c$. This proves <connected triple supports generate an alternating group>, by successively merging the intersecting triples of a connected support hypergraph. We used that <three-cycles> generate <alternating groups>: pairs of <transpositions> with a common letter are <three-cycles>, while a pair of disjoint <transpositions> is a product of two <three-cycles>.
Now take all conjugates in $G$ of the given <three-cycle>. Their support hypergraph is $G$-invariant, and its connected components form a <block system>. Transitivity makes their union the whole set. There is an edge, so a component has more than one vertex. A <primitive group action> forces a single component. The preceding merging argument proves
$$
\boxed{A_n\le G.}
$$
For the first family of counterexamples, let
$$
G=PGL_2(p)\curvearrowright\mathbb P^1(\mathbb F_p),\qquad p>3.
$$
Its elements are <Möbius transformations> $x\mapsto(ax+b)/(cx+d)$, modulo nonzero scalar <matrices>. Any ordered triple of distinct <projective points> is the image of $(\infty,0,1)$ under exactly one such transformation: representatives of the first two target lines form a <basis>, and scaling the two columns sends their sum to the third target line. Thus the action is <sharply three-transitive>, hence <two-transitive> and primitive. Its degree is $p+1$ and order $p(p^2-1)$. It is divisible by $p$, but
$$
\frac{|A_{p+1}|}{|PGL_2(p)|}=\frac{(p-2)!}{2}>1,
$$
so it cannot contain $A_{p+1}$.
For the second example take
$$
\boxed{p=7,\quad G=PGL_2(8),\quad n=9,\quad |G|=504.}
$$
The same projective-line construction over $\mathbb F_8$ is <sharply three-transitive>. Its order is divisible by 7 and is less than $|A_9|=181440$, so it does not contain $A_9$. For a concrete field one can use $\mathbb F_8=\mathbb F_2[u]/(u^3+u+1)$.
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