---
tag: 78R2TV41D4
kind: statement
role: theorem
name: Jordan's theorem on primitive groups containing a prime cycle
msc: [20B15, 20B20, 20B30]
uses: [Y10S0751KG, 07CT9RDY7Q, 01826KMG2H, 0515A6D2Y1, 0Z5905Q73D, BDJ1AX4XBB]
about: [Y10S0751KG, 0515A6D2Y1, 0Z5905Q73D]
sources:
  - url: https://en.wikipedia.org/wiki/Jordan's_theorem_(symmetric_group)
same_as:
  wikidata: Q6276298
  wikipedia: Jordan's theorem (symmetric group)
---

## Statement

Let $n$ be a positive integer, and let $G$ be a [[primitive permutation group|Y10S0751KG]] on a set $\Omega$ with $n$ elements.
Let $p$ be a [[prime number|01826KMG2H]] with $p \le n - 3$.
Suppose that $G$ contains a [[cycle|07CT9RDY7Q]] of length $p$.
Then $G$ is the [[alternating group|0Z5905Q73D]] $A_n$ or the [[symmetric group|0515A6D2Y1]] $S_n$ on $\Omega$.

## Remarks

The proof uses that the $p$-cycle fixes at least three points.
Combining this with primitivity, one shows that $G$ is [[multiply transitive|BDJ1AX4XBB]], 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 $n - 3$.

## Explanations

### L1 human

Primitive groups shuffle a set without respecting any grouping of its points.
Jordan's theorem says that such a group is very large as soon as it contains one special element.
This element is a cycle of prime length that leaves at least three points in place.
Then the group contains all even permutations.
So a single element, together with primitivity, can force a group to be huge.

### L2 human

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|BDJ1AX4XBB]] and the sets of fixed points of elements.