# Is S_6 the automorphism group of a group?

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?

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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 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.

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The link is dead. A working one is dx.doi.org/10.1216/RMJ-1979-9-4-653 . –  Emil Jeřábek Apr 15 '11 at 16:49

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.

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There's a stronger result <a href="sciencedirect.com/…; for countable groups. –  Qiaochu Yuan Oct 19 '09 at 19:15