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

Any automorphism of a group $G$ is a bijection that fixes the identity, so an easy upper bound for the size of $\operatorname{Aut}(G)$ for a finite group $G$ is given by

\begin{align*}|\operatorname{Aut}(G)| \leq (|G| - 1)! \end{align*}

This inequality is an equality for cyclic groups of orders $1$, $2$ and $3$ and also the Klein four-group $\mathbb{Z}_2 \times \mathbb{Z_2}$. I think it's reasonable to believe that they are the only groups with this property. The factorial $(|G| - 1)!$ is eventually huge. I searched through groups of order less than $100$ with GAP and found no other examples.

The problem can be reduced to the abelian case. We can check the groups of order $< 6$ by hand. Then if $|G| \geq 6$ and the equality holds, we have $\operatorname{Aut}(G) \cong S_{|G|-1}$. Now $\operatorname{Inn}(G)$ is a normal subgroup of $\operatorname{Aut(G)}$, and is thus isomorphic to $\{(1)\}$, $A_{|G|-1}$ or $S_{|G|-1}$. This is because $A_n$ is the only proper nontrivial normal subgroup of $S_n$ when $n \geq 5$. We can see that $(|G| - 1)!/2 > |G|$ and thus $\operatorname{Inn}(G) \cong G/Z(G)$ is trivial.

How to prove that there are no other groups for which the equality $|\operatorname{Aut}(G)| = (|G| - 1)!$ holds? Are any better upper bounds known for larger groups?

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If $G$ is abelian, then inversion is a central automorphism. What does the center of the symmetric groups look like? – user641 Feb 13 '12 at 16:47
Since I added a different proof below, I wanted to finish the proof I started in the comment above: since (most of) the symmetric groups have trivial center, inversion is trivial on $G$. This means every element of $G$ is order $2$, and $G$ is an elementary abelian $2$-group. But it's easy to see that $|GL(n,2)| < (2^n-1)!$ for $n>2$; indeed, the $2^n-3$ factor is missing from the left side. – user641 Feb 13 '12 at 17:25

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 $|G| = |G_r| \geq 2^r.$

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The Klein 4 case still allowed equality, which is noted in the question. This is consistent with the argument above, but was obscured by the original wording, so I have done a minor edit. – Geoff Robinson Feb 13 '12 at 17:52
I had lectures with Neumann! Very thorough and witty, unfortunately they only gave him a 4 week module. – Adam Feb 13 '12 at 19:04
Although "better upper bounds" was a secondary question, I decided to accept this answer because I like that proof. Thanks to everyone else for answering too. – Mikko Korhonen Feb 17 '12 at 19:51
I've never seen that argument about the size of a generating set before - very cute! It seems like you could show the bound even more simply by noting that all of the products $\Pi_{i\in S}x_i$ for the $2^r$ subsets $S \subseteq \{1\ldots r\}$ are distinct (by minimality)? – Steven Stadnicki Feb 23 '12 at 18:12
@Steven Stadnicki : Maybe, but are you not assuming that the generators commute o even make sense of those products? I nearly put in the remark that using the minimality, we have $|{\rm Aut}(G)| \leq \prod_{j=1}^{r} \left( |G|-|G_{j-1}| \right)$ because for any automorphism $\alpha,$ we must have $\alpha(x_j)$ outside $\alpha(G_{j-1})$, since $\{ \alpha(x_1), \ldots \alpha(x_r) \}$ is also a minimal generating set. – Geoff Robinson Feb 23 '12 at 18:25

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.

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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 $x \neq ab$.

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Looks good to me. Very simple. – Jack Schmidt Feb 24 '12 at 14:24
@JackSchmidt: Cool, thanks. – Mikko Korhonen Feb 24 '12 at 14:50

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.

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Well, for starters, to get equality you would need every element to have the same order, and so that order is a prime (so your group has prime-power order).

You should next realise that if you have equality then you must have equality in $C_p$, the cyclic group of order $p$, as there must exist a homomorphism which switches every element of $\langle g\rangle$, where $g$ is an element of order $p$, and keeps every other element in you group fixed. Clearly this doesn't happen if $p=2$ or $p=3$.

For $p=2$, notice that is doesn't work for $C_2\times C_2\times C_2$, but that the observation in the above paragraph still must hold (that you can restrict to subgroups and still get an automorphism).

For $p=3$, notice that is doesn't work for $C_3\times C_3$, but that the observation in the above paragraph still must hold (that you can restrict to subgroups and still get an automorphism).

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