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If we have an algebra A over a finite group G, then if G is non-abelian we can have a non-trivial set of compact automorphisms of A that map the elements of G onto a set isomorphic to G. It may be that it suffices for this to contain the identity, but I haven't verified this.

For example, the algebra of the field $C$ over $S_3$ has such a set of automorphisms that form a group isomorphic to $SO(3)$. When G is dihedral of size $2p$, the A(G) have compact automorphisms satisfying the above requirement with at most $3(p-1)/2$ generators is $p$ is odd, or $3(p-2)/2$ if $p$ is even.

Has anyone classified these?

Many thanks to Nik Weaver for explaining the proper language with which to ask the question.

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    $\begingroup$ What's an "algebra over a group"? what's a "compact automorphism"? $\endgroup$
    – YCor
    Commented Aug 16, 2019 at 5:25
  • $\begingroup$ And what do you mean by "a set isomorphic to $G$"? $\endgroup$
    – Alex B.
    Commented Aug 16, 2019 at 12:02
  • $\begingroup$ I think of an element of Aut(A) as a transformation over the elements of A. N of the elements of A are the N elements of G, multiplied by the field element 1. This set, when its members are operated on by one of the automorphisms of interest here, produces a new set. The natural operations between the members of the new set are isomorphic to the operations between the original elements of G. One consequence is that the overall absolute value of any scaling is 1. $\endgroup$
    – James Bellinger
    Commented Aug 17, 2019 at 16:44

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