In a [recent question on MSE](https://math.stackexchange.com/questions/3537191/inner-automorphisms-of-group-algebras-vs-inner-automorphisms-of-the-group) I asked about conditions under which the canonical morphism $Out(G) \to Out(k[G])$ is injective.



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> Is it true that this morphism is indeed injective if $G$ is finite and $k=\mathbb{Z}$ ?
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What I already know:


* It's not injective if $k$ is a field in non-modular characteristic. The kernel is $Out\_c(G)$, the group of conjugacy class preserving automorphisms. That is [Captain Lama's answer](https://math.stackexchange.com/a/3537243/62443) to my question.
* It is true if $k=\mathbb{Z}$ and $G$ is nilpotent. That's [my own answer](https://math.stackexchange.com/a/3537416/62443) to the question.
* In the mean time I have also found out that the morphism is injective if $G$ is rational, i.e. if $g$ is cojugated to $g^k$ for every $g\in G$ and every $k\in\mathbb{Z}$ with $gcd(k,ord(g))=1$, e.g. $G$ a symmetric group.


Proof: If $\alpha\in Aut(G)$ is in the kernel of the morphism, say $\alpha(g)=ugu^{-1}$ for some $u\in\mathbb{Z}G^\times$, then
$$u^\ast u g=u^\ast \alpha(g) u = u^\ast \alpha(g^{-1})^{-1} u = (u^\ast \alpha(g^{-1}) u)^\ast = (u^\ast u g^{-1})^\ast = g u^\ast u$$
where $^\ast$ is the antiautomorphism of the group algebra with $g\mapsto g^{-1}$. Hence $u^\ast u \in\mathbb{Z}G$. We want to prove that $u^\ast u = 1$.


Consider all the complex characters $\chi\in Irr(G)$ and their central characters $\omega\_\chi: Z(\mathbb{C}G) \to \mathbb{C}$. We can pick a matrix representations $\rho\_\chi: G\to GL\_n(\mathbb{C})$ affording $\chi$ with $\rho(G)\subseteq U\_n(\mathbb{C})$ so that $u^\ast u$ is mapped to some self-adjoint and positive definite matrix so that $\omega\_\chi(u^\ast u)\in\mathbb{R}\_{>0}$. Furthermore $u^\ast u\in Z(\mathbb{Z}G)$ is integral over $\mathbb{Z}$ so that $\omega\_\chi(u^\ast u)$ is an algebraic integer. Moreover it must be in $\omega\_\chi(Z(\mathbb{Q}G)) =\mathbb{Q}(\chi)$. Now if $G$ is rational, then all characters have rational values so that $\omega\_\chi(u^\ast u)$ is a real, positive, rational, invertible integer. In other words $\omega\_\chi(u^\ast u) = 1$, i.e $\rho\_\chi(u^\ast u) = 1\_{n\times n}$. Since $\chi$ was arbitrary, $u^\ast u=1$.


The only units of $\mathbb{Z}G$ with $u^\ast u=1$ are elements of the form $\pm g$ so that $\alpha\in Inn(G)$ as we wanted. QED.


Note that we can get the same conclusion under weaker conditions. For example if $\mathbb{Q}(\chi)=\mathbb{Q}(i)$, then the only real, positive, integral unit is also 1.



Inner automorphisms of group algebras vs. inner automorphisms of the group
The question for finite $G$ and $k = \mathbb{Z}$ is the **normalizer problem**, see [1, Section 1]. By a result of Jan Krempa, the kernel of the cannonical morphism is in that case always an elementary abelian $2$-group. As far as I know, there is basically only one example known where the kernel is non-trivial [1, Theorem A]. This example was constructed by Martin Hertweck and used to provide a counterexample to the isomorphism problem for integral group rings.


[1] Martin Hertweck; A counterexample to the isomorphism problem for integral group rings. *Ann. of Math.* (2) **154** (2001), no. 1, 115–138, MR1847590.


