Primitive permutation groups. Suppose that G is a primitive permutation group of degree n which contains a 3–cycle. I want to show that G contains the alternating group.
Here G being a primitive permutation group of degree n means that G is a subgroup of the symmetric group $S_n$ and that the only G-congruences on $\{1,\ldots ,n\}$ are trivial.
G-congruence is an equivalence relation $R$ on $\{1,\ldots ,n\}$ such that $aRb$ implies $agRbg$ $\forall g \in G$. And a trivial G-congruence means that the relation is either universal or equality.
 A: Let $G$ be a primitive permutation group of degree $n$.  Assume $G$ contains a 3-cycle, say $s=(123)$.  If $G$ also contains another 3-cycle $t$ which is not disjoint from (123), then either $t$ is something like (124) which has 2 elements in common with $s$, or else $t$ is something like (145) which has only 1 element in common with $s$.  In the former case, $\langle s,t \rangle$ is all of $A_4$ and therefore contains all 3-cycles involving $\{1,2,3,4\}$, and in the latter case, $\langle s,t \rangle$ is all of $A_5$ and therefore contains all 3-cycles involving $\{1,2,3,4,5\}$.
Therefore, if we define the relation on $\{1,2,\ldots,n\}$ by $$aRb {\mbox{ if either }} a=b {\mbox{ or there is a 3-cycle }} (a,b,c) \in G$$
then $R$ is a $G$-congruence.  By hypothesis, $R$ is not equality (because $G$ contains a 3-cycle), therefore, by primitivity, there is only a single equivalence class.  Therefore, $G$ contains all 3 cycles.  Since $A_n$ is generated by 3-cycles, we are done.
(Reference: This is a generalization of the argument for 2-cycles, where a primitive permutation group containing a transposition must be all of $S_n$.  See , for example, Theorem 8.17 in Isaacs' Finite Group Theory on page 241.)
As mentioned in the comments, a more general form of this result is Jordan's theorem, which says that a primitive permutation group of degree $n$ containing a $p$-cycle where $p$ is prime and $p \le n-3$, must contain $A_n$.
