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Suppose we have an abelian extension of Hopf algebras,

$$k \rightarrow k^G \rightarrow A \rightarrow kF \rightarrow k.$$

According to the general theory there is a left action of $F$ on $G$ and a $2$-cocycle $\sigma:F\times F \rightarrow k^G$ such that $A=k^G$ #${}_{\sigma} kF$ as algebras.

1)Is it true that $\sigma^n=\mathrm{id}$ for some $n\geq 1$? In other words can one choose $\sigma'$ in the same class with $\sigma$ such that if $\sigma'(f, h)=\sum_{a \in G}\sigma'_a(f,h)p_a$ then $$\sigma\;'_a(f,h)^n=\mathrm{id}$$ for all $a, f, h$. Here $p_a$ is the usual notation for dual basis of the group element basis .

2) A little more general question (that implies the first question) is if the set of all $2$-cocycles of $F$ with values in $k^G$ finite? Equivalently the question is if $H^2(F, k^G)$ is finite with $F$ acting on $k^G$ via the action of $F$ on $G$.

3)A related question for what groups $X$ with an $F$-action the group $H^2(F, X)$ is finite? For example $X=k^*$ gives the usual Schur multiplier.

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What are the various things in your extension? Is it an extension of modules? – David Roberts Jan 9 '11 at 20:59
No they are all Hopf algebras. $kF$ is a group algebra and $k^G$ is the dual of a group algebra. It should be thought as an analogue to an extension of groups. – Sebastian Burciu Jan 9 '11 at 21:36
I fixed up the TeX for you. – David Roberts Jan 9 '11 at 22:51
Thanks for taking care of the Tex! – Sebastian Burciu Jan 10 '11 at 11:41
up vote 1 down vote accepted


Suppose that $k$ is an algebraic closed field of characteristic zero.

Let $G$ be a finite group and $X$ a finite right $G$-set, so $k^X$ is left $G$-module. We want to see that $H^2(G,k^X)$ is a finite group. Let $X=\cup_{i=1}^n X_i$ where each $X_i$ is a transitive $G$-set, then as $G$-module $k^X= \bigoplus_{i=1}^n k^{X_i}$, and $H^2(G,k^X)= \bigoplus_{i=1}^n H^2(G,k^{X_i})$. Now by Shapiro's Lemma link text, $H^2(G,k^{X_i})\cong H^2(G_i,k^*)$, where $G_i$ is the stabilizer of an element $x_i \in X_i$ and $G_i$ acts trivially over $k^*$. Finally, since $k$ is algebraic closed, the Schur multiplers $H^2(G,k^{X_i})$ are finites.

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Cesar, thank you very much! I knew that should be true but couldn't find a proof written somewhere. – Sebastian Burciu Jan 10 '11 at 16:03

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