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Let $S_k \wr S_n$ be the wreath product of two symmetric groups (so $S_n$ acts on $(S_k)^n = S_k \times ... \times S_k$ by permuting the factors; we then take the semi-direct product).

What is $Aut(S_k \wr S_n)?$

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For $k=5$ and $k>6$, $(S_n \wr S_k)$ is complete. For $n=6$ there is an additional outer automorphism of order 2, which comes from the outer automorphism of $S_6$ acting in the same way on all of the $S_k$ direct factors of the base group. I haven't thought yet about $n<5$. –  Derek Holt Dec 8 '10 at 13:13
    
OK - third attempt. Why can't comments be edited? For $k=5$ and for $k>6$, `$S_k \wr S_n$' is complete. For $k=6$ there is an additional outer automorphism of order 2, which comes from the outer automorphism of $S_6$ acting in the same way on each of the $n$ direct factors of the base group. I haven't thought yet about $k<5$. After a few computer calculations, it looks as though it is always complete for $k=4$ and for $k=3$ and $n>2$, but not for $k=3,n=2$. –  Derek Holt Dec 8 '10 at 13:39
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The usual workaround for editing comments is deleting the old one and replacing it with a new one in a short period of time. –  j.c. Dec 8 '10 at 15:27
    
@Holt: Arggg, your comments are all mashed-up with the TIPS box and are unreadable. Too bad there's no edit feature... –  Dr Shello Dec 8 '10 at 17:13
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Why don't you accept an answer if it answers your question? All your questions have answers and none are accepted. –  Alex B. Dec 15 '10 at 8:26

1 Answer 1

Let $G = S_k \wr S_n$. For $k \ge 5$, $G$ has the unique minimal normal subgroup $A_k^n$, which is therefore fixed by every automorphism $\phi$ of $G$. If $\phi$ centralizes $A_k^n$ then, for any $g \in G$, $h \in A_k^n$, we have $g^{-1}hg = \phi(g^{-1}hg) = \phi(g)^{-1}h\phi(g)$. So $\phi(g)g^{-1} \in G$ centralizes $A_5^n$, and hence $\phi(g)=g$, so $\phi=1$. Hence ${\rm Aut}(G) \le {\rm Aut}(A_k^n)$.

For $k \ne 6$, we have ${\rm Aut}(A_k^n) \cong G$, so $G$ is complete. If $k=6$, then it is not hard to see that $|{\rm Out}(G)| = 2$, where the outer automorphism of order 2 acts as the same outer automorphism of $S_6$ on each of $n$ factors of $S_6^n$.

I believe that $G$ is also complete when $k=4$ and when $k=3$ and $n>2$, although I have not tried to write down a proof. For $k=3$ and $n=2$ there is an outer automorphism of order 2.

I have not thought at all about $k=2$, but note that $G$ has centre of order 2 in that case.

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Oh, this is very nice; thank you. Do you know a reference for the irreducible representations of such groups being worked out explicitly? –  Dr Shello Dec 10 '10 at 4:01

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