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Let $k$ be a commutative ring, $G$ a finite group and $\alpha\in\operatorname{Aut}(k[G])$ an automorphism of $k$-algebras.

If we know that $\alpha\in\operatorname{Inn}(k[G])$ and $\alpha(G)=G$, can we conclude that there is a group element $g\in G$ with $\alpha(x)=gxg^{-1}$ ? In other words: Is the canonical map $\operatorname{Out}(G) \to \operatorname{Out}(k[G])$ injective?

Such an $\alpha$ looks suspiciously like an inner automorphism of $G$. For example, $\alpha$ maps every conjugacy class to itself. In particular it acts trivially on $Z(G)$ and every normal subgroup is $\alpha$-invariant. But is it really an inner automorphism of $G$?

If it is not generally true, is it at least true for some special rings like $k=\mathbb{Z}$ for example?

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  • $\begingroup$ At least when $k$ is a field of characteristic $0$ (or coprime with $|G|$) this should be equivalent to: "If an automorphism of $G$ acts trivially on conjugacy classes, is it inner?". $\endgroup$ Feb 6 '20 at 21:54
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When $k$ is a field of characteristic coprime with $|G|$ (so for instance characteristic $0$), then $k[G]$ is semi-simple, so any automorphism fixing the center is inner by Skolem-Noether theorem. Since the center is generated by elements $\sum_{g\in C}g$ for each conjugacy class $C$, this means that the condition on $\alpha\in \operatorname{Aut}(G)$ is exactly that it acts trivially on conjugacy classes.

Now this article https://arxiv.org/pdf/1002.1359.pdf shows that there are such automorphisms which are not inner.

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  • $\begingroup$ You need the field to be a splitting field for $G$ for Skolem-Noether to apply, otherwise the Wedderburn components might not be central simple algebras. Apart from that, you make a very good point. I didn't realise that. But the second question still stands: Can we get a positive answer by choosing a $k$ which is not a field, say $k=\mathbb{Z}$ ? $\endgroup$ Feb 6 '20 at 22:59
  • $\begingroup$ They will be central simple algebras over separable extensions of $k$, which is enough. $\endgroup$ Feb 6 '20 at 23:02
  • $\begingroup$ If I've not made a mistake, I think $k=\mathbb{F}_p$ works when $G$ is the group of $3\times 3$ upper triangular, unipotent matrices with entries from $\mathbb{F}_q$ and $q=p^r$ with $r>1$. This group has outer automorphisms that preserve all conjugacy classes, but one can write them all down and stare at it for a while and (if I've not stared wrong) see that none of them come from an inner automorphism of $\mathbb{F}_p[G]$. $\endgroup$ Feb 7 '20 at 1:05
  • $\begingroup$ It's possible (hard to say without seeing the example), but this is not a contradiction, since you are taking a $p$-group and a field of characteristic $p$. $\endgroup$ Feb 7 '20 at 6:46
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$\newcommand{\IF}{\mathbb{F}}$ Here is a partial result for nilpotent groups:

Lemma: If $G$ is a $p$-group and $\IF$ a field of characteristic $p$, then $Out(G) \to Out(\IF[G])$ is injective.

Proof: Let $\alpha$ be conjugation with $u\in\IF[G]^\times$ such that $\alpha(G)=G$.

Write $u=\sum_{x\in G} \lambda_x x$ with $\lambda_x\in\IF$. Then $\forall g: ug=\alpha(g)u$ is equivalent to $\forall g,x: \lambda_{\alpha(g)xg^{-1}} = \lambda_{x}$.

Now consider the action of $G$ on $G$ via ${^g x}:=\alpha(g)xg^{-1}$ and the augmentation map $\nu:\IF[G]\to\IF$. The map $x\mapsto\lambda_x$ is constant on $G$-orbits w.r.t. to this map so that: $$0\neq \nu(u) = \sum_{x\in G} \lambda_x = \sum_{\substack{x\in G \\ |^G x|=1}} \lambda_x$$ because $char(\IF)=p$ and all orbits have a $p$-power length. In particular: There must be at least one $x\in G$ that is fixed under this action, i.e. $\forall g\in G: \alpha(g)xg^{-1} = x$ which means $\forall g: \alpha(g)=xgx^{-1}$ which we wanted to prove.

Corollary: If $G$ is nilpotent, then $Out(G)\to Out(\mathbb{Z}[G])$ is injective.

A nilpotent group is the product of its sylow subgroups $G=G_{p_1}\times G_{p_2}\times...\times G_{p_m}$. We induce over $m$. For $m=1$ we use the lemma.

For the induction step consider more generally $G=G_1\times G_2$. Then the two projections $G\to G_i$ induce automorphisms $\alpha_i\in Inn(\mathbb{Z}[G_i])$. By induction we can assume that there exists group elements $x_i\in G_i$ such that $\forall g_i\in G_i: \alpha_i(g_i) = x_i g_i x_i^{-1}$. Since $G$ is the direct product, $\alpha$ is conjugation by $x=(x_1,x_2)$.

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