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In the book Twelve sporadic groups, Griess states

If $A$ is an abelian group, $G$ acts on $A$, $z\in Z(G)$ satisfies $z-1\in$ Aut$(G)$, then $H^n(G,A)=0$ for $n\geq 0$. This is an observation of Jack McLaughlin.

My first simple question is what do we mean here by $z-1\in$ Aut$(G)$?

Second question, I didn't get reference of it in google search; can one state a reference for a modern, elementary proof of it? (I mean, if a proof of this fact appeared in some book or recent papers, please state it.)

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    $\begingroup$ Is it a typo for $\operatorname{Aut}(A)$? I'd expect $z-1$ to mean $a \mapsto (z-1)a$: if this was invertable it would certainly imply $H^0(G,A)=0$. $\endgroup$
    – M T
    Commented Jan 30, 2017 at 15:13
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    $\begingroup$ Using the finite-groups tag is confusing, because there are no finiteness assumptions in this question. $\endgroup$
    – Derek Holt
    Commented Jan 30, 2017 at 22:13

1 Answer 1

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The statement is not identical, but it is the "center kills" argument, as in 149 page 42 of these lecture notes.

For $g \in Z(G)$, the map $m \mapsto gm$ is a ${\mathbb Z}G$-automorphism of $A$. The map it induces on the cohomology groups $H^n(G,A)$ is the so-called conjugation map $c_g$, and since $g \in G$, this is the identity map on $H^n(G,A)$. Of course, the same map with $g=1$, $m \mapsto m$ also induces the idenity on $H^n(G,A)$, so $g-1$ induces the zero map.

But if $g-1$ is an automorphism of $A$, then it is a ${\mathbb Z}G$-automorphism, and induces an automorphism of $H^n(G,A)$. So $H^n(G,A)=0$.

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