# Intuition for cohomology of groups [duplicate]

• This is not intuition, but possible motivation : if you are interested in computing the rational points of an elliptic curve $E$, say $y^2 = x^3 - k$, then you notice that $E(\Bbb Q)$ is the set of invariants of $E(\overline{\Bbb Q})$ under the action of the absolute Galois group. Taking these invariants provides a left-exact functor [unfortunately it is not right exact, i.e. invariants don't commute with quotients], whose right derived functors are precisely giving group cohomology (here Galois cohomology). – Watson Jun 2 '18 at 14:30
• Dear @Pierre21, the equation $y^2 = x^3 - k$ (for a fixed non-zero integer $k$) is the equation of an elliptic curve, and $E(\Bbb Q)$ is the set of pairs $(x,y)$ satisfying this equation (there is just a "subtelty" with the "point at infinity"). This is not hard at all! (What is harder is to have a group structure on $E(\Bbb Q)$, though). – Watson Jun 2 '18 at 14:46