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I am looking for an elementary proof that the cohomology groups in the title are trivial in the positive degrees.

In more detain, let $G=\{1,s\}$ be a group of order two, and let $A$ be an abelian group with an action of $G$. I need an elementary proof that if $A$ is finite of odd order, then $H^1(G,A)=0$ and $H^2(G,A)=0$.

A non-elementary proof goes as follows. Since $|G|=2$, both $H^1(G,A)$ and $H^2(G,A)$ are killed by the multiplication by $2$. Clearly they are also killed by the multiplication by $|A|$. Since $2$ and $|A|$ are coprime, these cohomology groups are killed by the multiplication by $1$, hence they both are trivial.

I give an elementary formulation of the desired assertion. Let $A$ be an abelian group and let $s\colon A\to A$ be an automorphism such that $s^2=1$. Set $$ N=s+1, \qquad T=s-1. $$ Then $TS=0$ and $ST=0$, hence $$ \mathrm{im\ } N\subseteq \ker T\quad\text{and}\quad \mathrm{im\ } T\subseteq \ker N. $$ I need an elementary proof of the following assertion:

Theorem. If $A$ is a finite abelian group of odd order, then $\ker T=\mathrm{im\ } N$ and $\ker N=\mathrm{im\ } T$.

Motivation. Let $X$ be a quasiprojective variety with additional structure over ${\mathbb{C}}$. Write $G=\mathrm{Gal}({\mathbb{C}}/{\mathbb{R}})$, then $G=\{1,s\}$, where $s$ is the complex conjugation. Let $sX$ denote the variety with additional structure over ${\mathbb{C}}$ obtained from $X$ by the action of the complex conjugation $s$ on the coefficients of the equations defining $X$. Assume that $sX$ is isomorphic to $X$. We would like to know whether $X$ admits a real form. Let $A=\mathrm{Aut}(X)$, and assume that $A$ is an abelian group. Then one can construct an obstruction $\eta(X)\in H^2(G,A)$ to the existence of a real form of $X$, see my question. Now assume that $A$ is a finite abelian group of odd order. Then by our theorem we have $H^2(G,A)=0$, hence $\eta(X)=0$ and therefore, $X$ admits a real form $X_{\mathbb{R}}$. The set of isomorphism classes of such real forms is a principal homogeneous space of the abelian group $H^1(G,A)$, see again my question. By our theorem we have $H^1(G,A)=0$, hence this real form is unique.

Motivation for an elementary proof: A potential reader of my paper comes from analysis and would prefer an elementary proof.

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Suppose that $x \in \ker T$, so $s(x)=x$. Since $x$ has odd order, there exists a positive integer $n$ (in fact $n= (o(x)+1)/2$) with $2nx=x$. Then, since $s(nx)=nx$, we have $x = s(nx) + nx$, so $x \in {\rm im}\ N$.

$\ker N \subseteq {\rm im\ } T$ follows by a similar argument, or by comparing orders.

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  • $\begingroup$ Thank you Derek, this is exactly what I need for my paper! $\endgroup$ – Mikhail Borovoi Jul 4 '16 at 16:13

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