## Group cohomology

Last update: 01 February 2012

## Group cohomology

Let $G$ be a group and let $M$ be an abelian group with an action of $G$ by automorphisms, i.e. let $M$ be a $G-$module. Define abelian groups where the operation in ${C}^{n}\left(G,M\right)$ is given by $\left({f}_{1}+{f}_{2}\right)\left({g}_{1},...,{g}_{n}\right)={f}_{1}\left({g}_{1},...,{g}_{n}\right)+{f}_{2}\left({g}_{1},...,{g}_{n}\right).$ Define $0→ C0GM →∂1 C1M →∂2 C2M →∂3 ⋯ such that ∂n+1 ∂n=0,$ by defining $(∂nf) g1...gn = g1f g2...gn + ∑i=1n -1 i f g1 ... gi-1 gigi+1 gi+2 ... gn + -1 n f g1... gn-1 ,$ for all $f\in {C}^{n-1}\left(G,M\right)$ and ${g}_{1},...,{g}_{n} \in G.$

• The $n-$cochains are the elements of ${C}^{n}\left(G,M\right).$
• The $n-$coboundary map is ${\partial }_{n}:{C}^{n}\left(G,M\right)\to {C}^{n+1}\left(G,M\right).$
• The $n-$cocycles are the elements of $\mathrm{ker}{\partial }_{n+1}.$
• The $n-$coboundaries are the elements of $\mathrm{im}{\partial }_{n}.$
• The ${n}^{th}$ cohomology group of $G$ with coefficients in $M$ is the abelian group $HnGM = ker∂n+1 im∂n .$

Let $𝔼$ be a field and let $G$ be a finite subgroup of $\mathrm{Aut}\left(𝔼\right)$.

1. ${H}^{1}\left(G,𝔼\right)$ is trivial.
2. ${H}^{1}\left(G,{𝔼}^{*}\right)$ is trivial.

 Proof. b. Let $f:G\to {E}^{*}$ be a cocycle so that By Dedekind's lemma we may choose $x\in 𝔼$ such that $b= ∑ g∈G fg gx$ is nonzero. If $h\in G$ then $hb= ∑ g∈G (hfg) ⋅hgx = ∑ g∈G f h - 1 f(hg)⋅ hgx= f h - 1 b.$ So $f\left(h\right)=b/hb$ and $f$ is a coboundary. $\square$

(Dedekind's Lemma.) Let $𝔽$ be a subfield of $𝔼$.

1. Distinct embeddings of $𝔽$ into $𝔼$ are linearly independent.
2. Distinct characters $\chi :G\to {𝔼}^{*}$ are linearly independent.
3. Distinct elements of $\mathrm{Aut}\left(𝔼\right)$ are linearly independent in ${\mathrm{End}}_{𝔽}\left(𝔼\right)$.

 Proof. Let $𝔼$ be a finite extension of $𝔽$. Let be algebra homomorphisms. Let ${c}_{1},...,{c}_{n} \in 𝔼$ be such that $∑i=1n cigi =0.$ For any $x,y\in A,$ $0= ∑i=1n cigi (xy) -gn(x) ∑i=1n cigi (y) = ∑i=1 n-1 ci( gi(x) -gn(x) ) gi(y).$ So, by induction, For each $1\le i\le n-1,$ we may choose $x$ such that ${g}_{i}\left(x\right)\ne {g}_{n}\left(x\right),$ and conclude that So $∑i=1n ligi = lngn =0.$ Thus $0={l}_{n}{g}_{n}\left(1\right)={l}_{n}.$ $\square$

## Notes and References

Where are these from?

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