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Let $G$ be a finite group, $M$ a trivial $G$-module and let $f:G \times G \to M$ be a 2-cocycle.

Question: Are the following maps $f_1,f_2$ group homomorphisms ?

$$f_1: G \to M,\; g \mapsto \sum_{i=1}^{|G|}f(g,g^i)$$ $$f_2: G \to M,\; g \mapsto \sum_{h\in G}f(g,h)$$

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Please indicate you have made some effort to solve the problem yourself. It's usually better to ask a question that helps you to understand the problem, rather than asking the problem itself. – Ben Millwood Feb 25 '13 at 0:38
At the moment, it's not clear you even understand the problem, so one can only guess how to answer it in a way that's useful. Do you know what a cocycle is? etc. – Ben Millwood Feb 25 '13 at 0:40
Ben, I posted a well-formulated question. A qualified answer gives either a proof that $f_i$ is a group homomorphism or provides a counterexample. I feel statements like "it's not clear you even understand the problem" or "do you know what a cocycle is ?" humiliating. Please be polite. Thanks. – user63850 Feb 25 '13 at 1:01
Well @user63850, you may have a point there, yet if you'd add some explanation, ideas, insights to your question, which is not basic mathematics and thus, one assumes, is asked by at least an advanced undergraduated, no one will be able to ask you those questions... – DonAntonio Feb 25 '13 at 1:03
I'm sorry, my intention was not to humiliate. But I feel like you're expecting more effort on the part of the answerer than you've provided yourself, and I don't think that's fair. I have no trouble believing that you understand the problem – you might be extremely intelligent. But you've not shown that, so I don't know what it is that you need explaining. If you'd spend a bit more time explaining why you are asking this question and why you can't answer it yourself, it would encourage much better answers. – Ben Millwood Feb 25 '13 at 1:06
up vote 1 down vote accepted

Of course, no. Take an arbitrary map $\alpha:G\to M$. Then $f(g,h)=\alpha(g)-\alpha(gh)+\alpha(h)$ is a cocycle which does not satisfy your conditions.

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Good answer. Thank you. – user63850 Feb 28 '13 at 10:07

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