0

In a set of $n$ elements the number of bijective mappings is $n!$. In a group $G$ with $n$ elements, the number of automorphism is less than $n!$, because we require that any automorphism $f:G\to G$ to to satify $f(e) = e$, being $e$ the neutral element of the group. Thus we have that:

$$|\text{Aut}(G)| \le (n-1)!$$

However, almost certainly that upper bound is too high and can be improved. How can we give a better bound on the number of automorphisms of a general finite group?

Shaun
  • 44,997
Davius
  • 853
  • 3
    But with $m=n!$ for $S_n$, we have $$m=|{\rm Aut}(S_n)|=|S_n|\color{red}{\le}(m-1)!,$$ @Michael. – Shaun Sep 01 '22 at 14:25
  • 1
    Assuming we have a set of generators for $G$, an automorphism of $G$ is determined by where it sends each generator. If there are $k$ generators, then there are at most $(n-1)!/(n-1-k)!<(n-1)^k$ ways to pick where each generator goes (given that two generators cannot go to the same element). While not all of these will be automorphisms, this is at least an upper bound. I suspect that we can bound the number of generators above by $\log_2(n)$, although I do not know this for sure. – Aaron Sep 01 '22 at 14:29
  • @Aaron Yes that's right, we have $|{\rm Aut}(G)| \le n^{\log_2 n}$, which is close to best possible. – Derek Holt Sep 01 '22 at 14:30

0 Answers0