MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The automorphism group of the symmetric group $S_n$ is $S_n$ when $n$ is not $2$ or $6$, in which cases it is respectively $1$ and the semidirect product of $S_6$ with the (cyclic) group of order $2$. (For this famous outer automorphism, see for instance wikipedia or Baez's thoughts on the number $6$.)

On the other hand, $S_2$ is the automorphism group of $Z_3$, $Z_4$ and $Z_6$ (and only those groups among finite groups). Hence my question: is $S_6$ the automorphism group of a group? of a finite group?

share|cite|improve this question
up vote 77 down vote accepted

${\rm S}_6$ is not the automorphism group of a finite group. See H.K. Iyer, On solving the equation Aut(X) = G, Rocky Mountain J. Math. 9 (1979), no. 4, 653--670, available online here.

This paper proves that for any finite group $G$, there are finitely many finite groups $X$ with ${\rm Aut}(X) = G$, and it explicitly solves the equation for some specific values of $G$. In particular, Theorem 4.4 gives the complete solution for $G$ a symmetric group, and when $n = 6$ there are no such $X$.

share|cite|improve this answer

It's probably worth pointing to

Belolipetsky, Mikhail; Lubotzky, Alexander. Finite groups and hyperbolic manifolds. Invent. Math. 162 (2005), no. 3, 459-472. MR2198218.

where it is shown that for every finite group G, there's an infinite group Gamma with Out(Gamma) = G.

share|cite|improve this answer
There's a stronger result <a href="…; for countable groups. – Qiaochu Yuan Oct 19 '09 at 19:15

On the other hand, ${\rm S}_6$ is isomorphic to ${\rm Sp}_4(\mathbb{F}_2)$, so that it is an automorphism group in another category (other than groups or sets). This automorphism is exhibited by looking at the 2-torsion of the Jacobian of a hyperelliptic curve H of genus 2 (if $H$ is given by $y^2 = f(x)$, with $f$ of degree 6, then 15 non-trivial two torsion points are given [as a Galois module] by differences of roots of $f$; see the wiki page for Kummer surface).

share|cite|improve this answer
You mean Sp_4(F_2), I am sure, - SL_4(F_2) is simply too big to be S_6. – Vladimir Dotsenko Nov 3 '09 at 22:59

There is a whole array of results, going back to G. Birkhoff at 1930s saying that every group is an automorphism group of some universal algebra (or some universal algebra inside some class).
(This really should be merely a comment to the previous answer, but I am still not reputable enough to leave comments).

share|cite|improve this answer
Meanwhile you have enough points to comment! – Stefan Kohl Dec 7 '15 at 17:52

Your Answer


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.