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.

  • 3
    $\begingroup$ Would you recall how Out of a $k$-algebra is defined? $\endgroup$
    – YCor
    Feb 22, 2020 at 22:31
  • 2
    $\begingroup$ 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 $u\in A^\times$. $\endgroup$ Feb 23, 2020 at 15:13

1 Answer 1


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.

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

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