From: "Dr. Robert Baddeley"Date: Mon, 17 Mar 1997 10:53:52 GMT0BST Subject: Re: Automorphisms of a wreath product Dear Jaoa, Let $G$ be a group, and let $P$ be a transitive permutation group on $\Omega$ (possibly infinite). Let $W$ be the restricted wreath product $G\wr P$. Denote the base group by $B$ (which is a direct product of copies of $G$). If $B$ is a characteristic subgroup of $W$, then we can view $Aut W$ as the normalizer in $Aut B$ of $W$. So we are interested in when $B$ is not characteristic. Fletcher Gross, in a paper `On the uniqueness of wreath products' J. Algebra (which has appeared but I'm not sure when), proved that if one of the following holds: (i) $P$ is not isomorphic to any proper factor group of itself; (ii) for some positive integer $n$ the $n^{th}$ term $G^{(n)}$ of the derived series of $G$ is not isomorphic to a proper subnormal subgroup of itself; (iii) the stabilizer in $P$ of an element of $\Omega$ has finite order; then $B$ is not characteristic subgroup only if $P$ has a non-identity normal elementary abelian 2-subgroup and $G=\langle u\rangle H$ where $H$ is abelian, $u^2=1$, and $u^{-1} h u = h^{-1}$ for all $h\in H$. Gross says more in the finite case. His results have been extended a little by Laci Kov\'acs and myself (as yet unpublished). If $W$ is finite wtih $P\le S_n$ and $B$ is not characteristic, then $G= \langle u\rangle H$ with $H$ abelian of odd order, $u^2=1$, $h^u=h^{-1}$ for all $h\in H$, and $P$ contains a normal elementary abelian 2-group of size $2^{n/2}$; moreover the $Aut W$ orbit of $B$ has length 2 precisely. In particular, if $P$ is the full symmetric group then $B$ is characteristic!! Cheers, Robert.