Every finite cyclic group occurs as the outer automorphism group of a finite simple non-abelian group

Statement

Let $n$ be any natural number. Then, there exists a Simple non-abelian group (?) $G$ such that the cyclic group of order $n$ occurs as the Outer automorphism group (?) of $G$.

Related facts

• Schreier conjecture
• Automorphism group of simple non-abelian group need not be ambivalent

Facts used

1. Projective special linear group is simple for any field of four or more elements.

Proof

The cases $n = 1$ and $n = 2$

For the case $n = 1$, there are plenty of examples, such as Mathieu group:M11.

For the case $n = 2$, there are again plenty of examples, such as the alternating group of degree five, whose automorphism group is the symmetric group of degree five and whose outer automorphism group is cyclic of order two.

The case $n \ge 3$

Suppose now that $n \ge 3$. Consider the group $PSL(2,2^n)$. This is clearly non-abelian; it is simple by fact (1). The outer automorphism group of $PSL(2,2^n)$ is the group of field automorphisms of the field with $2^n$ elements, which is cyclic of order $n$. This completes the proof.