Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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
2  
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
add comment

5 Answers

up vote 13 down vote accepted

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

share|improve this answer
    
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
show 2 more comments

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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 $x \neq ab$.

share|improve this answer
    
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
add comment

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).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.