Upper bounds on the size of $\operatorname{Aut}(G)$

Solution 1:

Even without the classification of finite simple groups, quite reasonable bounds are known, for example in work of P.M. Neumann. If the group $G$ can be generated by $r$ but no fewer elements, then no automorphism of $G$ can fix the $r$ given generators, so there are at most $\prod_{j=1}^{r} (|G|-j)$ different automorphisms of $G,$ since the $r$ generators must have distinct images, none of which is the identity. As P.M. Neumann has observed, $G$ can always by generated by ${\rm log}_{2}(|G|)$ or fewer elements, so we have $r \leq \lfloor {\rm log}_{2}(|G|) \rfloor .$ For $|G| >4,$ this always gives a strictly better bound for the size of ${\rm Aut}(G)$ than $(|G|-1)!.$ For large $|G|,$ it is much better. Using the classification of finite simple groups, much better bounds are known.

Later edit: Perhaps I could outline Neumann's argument, since it is quite elementary, and I do not remember a reference: Let $\{x_1, x_2, \ldots, x_r \}$ be a minimal generating set for $G$ and let $G_i = \langle x_1, x_2, \ldots, x_i \rangle $ for $i >0,$ $G_{0} = \{ e \}.$ Then for $1 \leq i \leq r,$ we have $|G_i| > |G_{i-1}|$ by minimality of the generating set. Furthermore, $|G_i|$ is divisible by $|G_{i-1}|$ by Lagrange's theorem, so $|G_i| \geq 2|G_{i-1}|.$ Hence $|G| = |G_r| \geq 2^r.$

Solution 2:

I believe this is an exercise in Wielandt's permutation groups book.

$\newcommand{\Aut}{\operatorname{Aut}}\newcommand{\Sym}{\operatorname{Sym}}\Aut(G) \leq \Sym(G\setminus\{1\})$ and so if $|\Aut(G)|=(|G|-1)!$, then $\Aut(G) = \Sym(G\setminus\{1\})$ acts $|G|-1$-transitively on the non-identity elements of G. This means the elements of G are indistinguishable. Heck even subsets of the same size (not containing the identity) are indistinguishable. I finish it below:

In particular, every non-identity element of G has the same order, p, and G has no proper, non-identity characteristic subgroups, like $Z(G)$, so G is an elementary abelian p-group. However, the automorphism group is $\newcommand{\GL}{\operatorname{GL}}\GL(n,p)$ which, for $p \geq 3, n\geq 2$, only acts at most $n-1$-transitively since it cannot send a basis to a non-basis. The solutions of $p^n-1 \leq n-1, p \geq 3, n \geq 2$ are quite few: none. Obviously $\GL(1,p)$ has order $p-1$ which is very rarely equal to $(p-1)!$, when $p=2, 3$. $\GL(n,2)$ still can only act $n$-transitively if $2^n-1 > n+1$, since once a basis's image is specified, the other points are determined, and the solutions of $2^n-1 \leq n+1$ are also limited: $n=1,2$. Thus the cyclic groups of order 1,2,3 and the Klein four group are the only examples.

Solution 3:

I think I have found a different solution. Is this correct?

Suppose that $G$ is a group and $|G| > 4$. To prove that $|\operatorname{Aut}(G)| < (|G| - 1)!$, it is enough to find a bijection $f: G \rightarrow G$ with $f(1) = 1$ that is not an automorphism.

First of all, let $a$ be some nonidentity element of $G$. Let $b \in G \backslash \{1, a, a^{-1}\}$ and $x \in G \backslash\{1,a,b, ab\}$. Define the map $f: G \rightarrow G$ by $f(ab) = x$, $f(x) = ab$ and $f(g) = g$ for rest of the elements in $G$. Then $f$ is a bijection that fixes $1$, but $f(ab) \neq f(a)f(b)$ because $x \neq ab$.

Solution 4:

Consider $G$ as a subgroup of $S_{|G|}=S$ via the regular representation. Then the normalizer of $G$ is the holomorph of $G$, and saying $|Aut(G)|=(|G|-1)!$ is the same as saying that $N_S(G)=S$. In other words, $G$ is a normal subgroup of order $n$ in $S_n$. This can only happen for $n<5$, and you get the following four groups: trivial, order 2, order 3, Klein 4.