Proving there is no non-abelian finite simple group of order a Fibonacci number

Solution 1:

The paper of Florian Luca that proves this result, and was mentioned in the comments, is as follows: Fibonacci numbers, Lucas numbers and orders of finite simple groups, Journal of Algebra, Number Theory and Applications, vol. 4, no. 1, 23--54 (2004).

Unfortunately, the paper is in an obscure journal, no version exists on the arXiv, and the journal website does not appear to be, how shall we say, very good.

I see some ideas with how to prove this myself, but would need to know more about Fibonacci numbers to do so. I satisfied myself with a proof for the alternating groups. (The sporadics can be done by inspection.)

By Carmichael's theorem, every $F_n$ is divisible by some prime not dividing any smaller $F_m$. Since $p$ divides $F_{p\pm 1}$, this primitive divisor must be at least $n-1$. But $|A_n|$ is divisible by exactly those primes at most $n$, and grows faster than $F_n$. In particular, if $|A_n|=F_m$, then $m>n$. Thus you only need to worry about the case $m=n+1$. And there is one: $n=3$, which of course does not yield a simple group.