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Let $A$ be a finite abelian group and let $G\subset \operatorname{Aut}A$, with $G$ acting from the left. $G$ acts in a natural way (on the left) on $A$'s character group $\hat A$ by $g\chi = \chi\circ g^{-1}$. My question is, if we know something about $G$'s action on $A$, can we say anything about its action on $\hat A$? More specifically:

(1) If $G$'s action on $A\setminus\{0\}$ is free, is its action on $\hat A\setminus\{0\}$ free? (2) $G$'s action on $A$ is faithful by construction; is its action on $\hat A$ faithful? (3) If its action on $A\setminus\{0\}$ is transitive, is its action on $\hat A\setminus\{0\}$ transitive?

My intuition says yes to all 3 questions, but I really do not see the argument so I have no idea whether to trust it. I think it is based on the fact that (a) $A$ and $\hat A$ are isomorphic (though not canonically), and (b) $A$ is canonically isomorphic to $\hat{\hat{A}}$, and the canonical isomorphism commutes with the action of $G$. But logically these don't imply anything. I can't extract any info from (a) unless one of the isomorphisms $A\rightarrow\hat A$ commutes with the action of $G$ on each, but usually none of them do. And (b) has nothing direct to say about $\hat A$ or $G$'s action on it.

It would be enough to conclude (1), (2) and (3) if I could guarantee that there is an isomorphism $A\rightarrow\hat A$ that commutes with the action of $G$ up to twisting $G$ by some automorphism. This seems potentially plausible to me but I don't see the argument; it's true in the one case I've worked out by hand ($A=C_7$, $G=C_6 = \operatorname{Aut} A$), but there are lots of special reasons why it should be true in that case.

Ben Blum-Smith
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    For what it's worth, with a generic finite group I don't think its conjugacy classeses and (complex) irreducible representations have the same symmetry type with respect to the group's outer automorphisms. (This is the nonabelian generalization of whether $A$ and its dual are isomorphic as $G$-modules.) – anon Sep 05 '15 at 00:45
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    I think the answer to all of your questions is yes. I believe that more generally the numbers of orbits of $G$ on $A$ and on $\hat{A}$ are equal (but not necessarily all having the same lengths), and that would imply all of your conjectures. But I can't remember the reference, and I don't have time to try and prove it myself right now. Unfortunately, it is often but not always true that there is an automorphism of $G$ mapping $A$ to $\hat{A}$. I found a counterexample with $A=C_{13}$ and $G = F_{42}$ the Frobenius group of degree $7$ and order $42$, which has no outer automorphisms. – Derek Holt Sep 05 '15 at 11:49
  • @Derek How would an automorphism of $G$ yield a mapping $A\to\widehat{A}$, and how can a group of order $42$ be a subgroup of ${\rm Aut}(C_{13})$ which has order $12$? – anon Sep 05 '15 at 14:57
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    You said yourself that the result would be correct if there was an isomorphism $A \to \hat{A}$ that commuted with an automorphism of $G$, and that is what I meant. But you are right that my counterexample doesn't work, because the action of $F_{42}$ on $C_{13}$ is not faithful. I still don't believe that this is true in general - I will try and think of another counterexample. In fact I got confused by your Question (2), because if $G \le {\rm Aut}(A)$ then its action on $A$ is faithful by definition. – Derek Holt Sep 05 '15 at 15:43
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    I can get a counterexample with faithful action of $G=F_{42}$ by letting $A$ be elementary abelian of order $13^7$, in which, as a $G$-module, $A = A_1\oplus A_2$ with $|A_1|=13^6$ and $G$ acting faithfully on $A_1$, and $|A_2|=13$ where $G$ acts with kernel of order $7$ on $A_2$. – Derek Holt Sep 05 '15 at 15:48
  • @Derek Brief terminology query. $G$ is a group, so an automorphism of it is a group isomorphism $G\to G$. An isomorphism $A\to\widehat{A}$ that commutes with the action of $G$ would be called $G$-equivariant. But since $G$ is a subset of ${\rm Aut}(A)$, its elements are automorphisms (of $A$), so "automorphism of $G$" could mean "element of $G$," but then an isomorphism $A\to\widehat{A}$ commuting with a single element of $G$ does not imply it commutes with every element of $G$ (i.e. commutes with the action of $G$, is $G$-equivariant). Is this right? (Will delete comments after verification.) – anon Sep 05 '15 at 16:26
  • I don't understand how "automorphism of $G$" could mean "element of $G$". – Derek Holt Sep 05 '15 at 19:50
  • @DerekHolt - I edited question 2 to acknowledge your point about $G\leq \operatorname{Aut}A$. – Ben Blum-Smith Sep 06 '15 at 05:55
  • @anon - I could be wrong but I think this is what Derek Holt means (just following up my lines of thought in the OP) by "there is an automorphism of $G$ mapping $A$ to $\hat A$" etc.: there is an isomorphism $\phi: A\rightarrow\hat A$ and an automorphism $\psi:G\rightarrow G$ such that $G$'s action on $A$ commutes with $\phi$ after twisting by $\psi$. In other words, for every $a\in A$ and $g\in G$, $\phi(ga) = \psi(g)(\phi(a))$. This is what I take Derek to be saying happens often but not always. – Ben Blum-Smith Sep 06 '15 at 06:00
  • @DerekHolt - I think I get what your example is doing: the action of $F_{42}$ on $C_{13}$ is really the action of $C_6$, and we just made it $F_{42}$ to eliminate the automorphism of $C_6$ we need to twist by to get the action of $C_6$ on $C_{13}$ to commute with an isomorphism $C_{13}\rightarrow\hat{C}_{13}$. I think you added the abelian group of order $13^6$ to the example just to make sure the action of $G$ on $A$ was faithful. Is that right? – Ben Blum-Smith Sep 06 '15 at 06:28
  • Yes that's exactly right! – Derek Holt Sep 06 '15 at 20:17

1 Answers1

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This is a partial answer, answering questions (1) and (2) but not (3). The methods are ad-hoc. I remain interested in a more unified approach, and of course I'd also like to know the answer to (3).

The answer to question (2) is yes. By basic representation theory, the characters of $A$ separate the elements of $A$. Thus an automorphism $g$ of $A$ can't fix every character without also fixing every element. Thus the action of $G$ on $\hat A$ is faithful.

The answer to question (1) is yes. I argue combinatorially:

We assume $G$'s action on $A\setminus\{0\}$ is free, which implies the same is true of the restriction of the action to any subgroup of $G$, and in particular of the cyclic subgroup generated by any element $g\in G$.

The statement that $G$'s action on $\hat A\setminus \{0\}$ is free is equivalent to the statement that for any nontrivial character $\chi\in \hat A$, $g\chi = \chi$ implies $g=1$. Now any nontrivial character $\chi$ is a homomorphism to a cyclic group of order $r\geq 2$. The fibers of this homomorphism all have cardinality $m = |A|/r$. One of them contains $0$ and thus it contains $m - 1$ nonidentity elements, and there is also at least one other fiber, with $m$ nonidentity elements.

Suppose $g$ satisfies $g\chi = \chi\circ g^{-1} = \chi$. Then $\chi\circ g^{-1}$ has the same fibers as $\chi$, and it follows that $g^{-1}$, and thus $g$, acts separately on each fiber. Thus each fiber is a union of orbits for the action of $\langle g\rangle$ on $A$.

As observed above, the action of $\langle g\rangle$ on $A\setminus\{0\}$ is free, which means that all orbits for $\langle g\rangle$'s action on $A\setminus\{0\}$ have the same length, namely the order of $g$ (let's call it $d$). Thus $m$ (the cardinality of the fibers not containing the identity, of which as noted above there is at least one) is a multiple of $d$. But $m-1$ is also a multiple of $d$, since this is the cardinality of the part of $A\setminus\{0\}$ sitting in the fiber containing $0$.

Thus $d\mid (m,m-1)$, i.e. $d=1$, i.e. $g$ is order $1$, i.e. $g=1$. This proves the action of $G$ on $\hat A$ is free.

Ben Blum-Smith
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