# Inner automorphisms of group algebras vs. inner automorphisms of the group

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?

• 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?". – Captain Lama Feb 6 '20 at 21:54

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.

• 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 $k=\mathbb{Z}$ ? – Johannes Hahn Feb 6 '20 at 22:59
• They will be central simple algebras over separable extensions of $k$, which is enough. – Captain Lama Feb 6 '20 at 23:02
• 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 $\mathbb{F}_p[G]$. – Johannes Hahn Feb 7 '20 at 1:05
• 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 $p$. – Captain Lama Feb 7 '20 at 6:46

$$\newcommand{\IF}{\mathbb{F}}$$ 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 $$x=(x_1,x_2)$$.