Jordan's theorem on primitive groups containing a prime cycle

Remarks

The proof uses that the $p$-cycle fixes at least three points. Combining this with primitivity, one shows that $G$ is multiply transitive, and then that $G$ contains $A_n$.

The bound $p \le n - 3$ is necessary. The cyclic group generated by $(1\,2\,3\,4\,5)$ is primitive on $5$ points and contains a $5$-cycle, so $p = n$, but it is neither $A_5$ nor $S_5$. The group $\mathrm{PGL}_2(\F_5)$ of order $120$ acts primitively on the $6$ points of the projective line over $\F_5$. There the map $x \mapsto x + 1$ is a $5$-cycle, so $p = n - 1$. But $120$ is neither $360$ nor $720$, so the group is neither $A_6$ nor $S_6$. The group $\mathrm{PSL}_2(\F_8)$ of order $504$ acts $2$-transitively, hence primitively, on the $9$ points of the projective line over $\F_8$. For a generator $a$ of $\F_8^\times$, the map $x \mapsto ax$ fixes $0$ and $\infty$ and is a $7$-cycle, so $p = n - 2$. But $504$ is smaller than $\abs{A_9}$, so the group is neither $A_9$ nor $S_9$.

The theorem gives a practical test for a permutation group of degree $n$ to contain $A_n$. It is enough that the group is primitive and contains one cycle of prime length at most $n - 3$.

Going deeper

For a student

The theorem is a classical tool to prove that a given permutation group is the full symmetric or alternating group. It is used, for example, to show that the Galois group of a polynomial is $S_n$. An irreducible polynomial gives a transitive Galois group, and one then looks for suitable cycles. Similar results say that a primitive group containing a transposition is $S_n$, and a primitive group containing a $3$-cycle contains $A_n$. The proof works with multiply transitive group actions and the sets of fixed points of elements.