Enjoy your holiday break!ġ2/14/21 - Lesson #8 on solving rational equations #1-5 PC 11 December 14thġ2/10/21 - Lesson #5 on multiplication of rational expressions #1-4 (a,c,e.),5-7 and Lesson #6 on dividing rational expressions #1-4 (a only),6-9 PC 11 December 10th Part 1ġ2/08/21 - Lesson #3 on adding and subtracting rational expressions #1-8 (a,c) and Lesson #4 questions #1-7 (a,c) PC 11 December 8th Part 1 PC 11 December 8th Part 2ġ2/07/21 - Lesson #1 on rational expressions #1-8 (a,c,e.) & Lesson #2 questions I refer you to pages 104-106 of Ken Brown's cohomology book for details.01/17/22 - Review and Test on Inequalitiesġ2/15/21 - No homework. The choice depends on your lift of $\rho$ to $\psi$ but then there is a map $Q\to Z(N)$ that will measure the difference of your choices and your $f$ will change by a coboundary, i.e., by an element of $B^2(Q,Z(N))$. The vanishing of the $3$-cocycle means exactly that you can find one such choice via the procedure I described above. So in fancy language, the set of functions $f$ meeting the associativity is an $H^2(Q,Z(N))$-torsor and you need to fix first some extension giving rise to $\rho$ before you can make explicit the bijection with $H^2(Q,Z(N))$. Then you can further multiply this $f$ satisfying the asociativity condition by $\omega$ on the correct side to get a new mapping $Q\times Q\to N$ that satisfies the associativity condition and that gives your multiplication. That is the vanishing of the 3-cocycle means that some $f$ exists satisfying the associativity condition. Now how does $\omega$ fit in? Because the $3$-cocycle coming from $\psi$ and $f$ as above is trivial, it is the coboundary of a $2$-cocycle $Q\times Q\to Z(N)$ and you can basically multiply by this $2$-cocycle (or maybe its inverse), to modify your $f$ to get one that meets the associativity criterion. According to Brown, the failure of $f$ to satisfy this formula is given by the 3-cocycle $Q\times Q\times Q\to Z(N)$ which measures the "difference" between the two sides (which one can check lies in $Z(N)$) and which coincides with the one you are referring to coming from $\rho$. The condition you need for this to be associative is that $f(q,q')f(qq',q'')=\psi(g)(f(q',q''))f(q,q'q'')$. Then you would like to define the product on $N\times Q$ by $(n,q)(n',q') = (n\psi(q)(n')f(q,q'),qq')$. $$\rho: Q\rightarrow \operatorname(N)$ has $\alpha(n)$ conjugation by $n$. According to Wikipedia and ncat lab general group extensions
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