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

In a recent question on MSE I asked about conditions under which the canonical morphism $$Out(G) \to Out(k[G])$$ is injective.

Is it true that this morphism is indeed injective if $$G$$ is finite and $$k=\mathbb{Z}$$ ?

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 to my question.
• It is true if $$k=\mathbb{Z}$$ and $$G$$ is nilpotent. That's my own answer 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.

• Would you recall how Out of a $k$-algebra is defined?
– YCor
Feb 22, 2020 at 22:31
• Just as you'd think: $Out(A) := Aut_{k\text{-Alg}}(A) / Inn(A)$ where $Inn(A)$ is the normal subgroup consisting of automorphisms that are of the form $a\mapsto uau^{-1}$ for some unit $u\in A^\times$. Feb 23, 2020 at 15:13

## 1 Answer

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.

• Thank you. That's very interesting. Mar 1, 2020 at 22:59