''Commutative'' 2-cocycles Let ba  $G$ an abelian group and $L$ is $G$-module.
If $f$ is a 2-cocycle in $Z(G,L)$, is it true that $f(g,h)=f(h,g)$ for all $g,h \in G$? Or even for $\bar{f} \in H^{2}(G,L)$ is $\overline{f(g,h)}=\overline{f(h,g)}$, here $\bar{f}$ is the cohomology class of $f$.
Thank you
 A: The answer is no, and counterexamples are easy to construct using the isomorphism from Schreier's theorem, which says that $H^2(G,A)$ is isomorphic as a pointed set $0\in H^2(G,A)$ is the point) to $EXT(G,A)$ which is the set of short exact sequences $1\to A\to\hat G\to G\to 1$ up to equivalence, with the point being the semi-direct product $\hat G=A\rtimes G$.
The construction of the cocylces involved is easy. Suppose you have a short exact sequence $1\to A\to\hat G\to G\to 1$ (i.e. identify $A$ as a normal subgroup of some $\widehat G$ so that $\widehat G/A\cong G$). Then pick a system of representatives $\widehat g\in G$ for every $g\in G$. Then $\widehat g\widehat h=f(g,h)\widehat{gh}$ defines a function $f\colon G^2\to A$. The condition that $f$ is a $2$-cocycle falls out from the equality $(\widehat g\widehat h)\widehat k=\widehat g(\widehat h\widehat k)$, $f$. 
If $f$ is "commutative", then $\widehat g\widehat h=f(h,g)\widehat{h,g}=f(g,h)\widehat{g,h}=\widehat g\widehat h$ where the middle equality is from commutativity of $G$. So all you need for a counterexample is to find a non-abelian group with quotient $G$, kernel $A$, and a system of representatives that do not commute with each other.
What is the smallest non-abelian group? $S^3$ of order $6$! So let $A=\left<(123)\right>\cong\mathbb Z/3$ and $G=S^3/A\cong\mathbb Z/2$. Then $(12)$ and $(123)$ are coset representatives of $A$ in $S^3$, but $(12)(123)=(13)$ while $(123)(12)=(23)$. So $f((123),(12))\neq f((12),(123))$.

What about the second question: if this is necessarily the case for the homology classes? Well, the different cocycles come from different choices of coset representatives, but the cohomology class of the cocycle is determined by the choice of short exact sequence $1\to A\to \widehat G\to G\to 1$. So an example when $\bar f(g,h)\neq\bar f(h,g)$ would be one in which there is no choice of representatives that commute with each other.
I suspect such an example is furnished by the next smallest non-abelian group: the quaternions of order $8$, which are $\{1,-1,i,-i,j,-j,k,-k\}$ with multiplication table $i^2=j^2=k^2=ijk=-1$, and the requirement that $A=\{1,-1\}$ is the center subgroup (then $G\cong\mathbb Z/2\times\mathbb Z/2$). It would be a fun exercise to think through the cocycles that various coset representatives give you.

P.S. Not every $2$-cocycle arises as a system of representatives in a short exact sequence! But every cohomology class is represented by such a cocycle in some (unique up to isomorphism) short exact sequence. That is the content of Schreier's theorem.
