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According to Silverman(Arithmetic of Elliptic curves):

The definition of a crossed homomorphisms, is a map $f : G \to M$ satisfying $f(ab)=bf(a)+f(b)$ for all $a$, $b$ in G.

According to many other books:

The definition of a crossed homomorphisms, is a map $f : G \to M$ satisfying $f(ab)=f(a)+af(b)$ for all $a$, $b$ in G.

Whether both the definitions are the same or different? If both the definitions are same then how I can define an action of $G/N$ on the first cohomology group $H^1(N,A)$.(where $N\trianglelefteq G$)

math
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    See (https://math.stackexchange.com/questions/2031287/induced-action-of-quotient-group-on-the-subgroup-cohomology) – Parthiv Basu Jul 27 '19 at 05:26
  • why $ (g\cdot f)$ is a crossed homomorphism w.r.t that action? @ParthivBasu – math Jul 27 '19 at 12:47
  • There is no trick to it, you just have to do an explicit computation to prove that $ (g \cdot f)(ab) = (g \cdot f)(a) + a (g \cdot f) (b) $. – Parthiv Basu Jul 27 '19 at 13:15
  • $(g\cdot f)(ab)=gf(g^{-1}abg)=gf(g^{-1}agg^{-1}bg)=gf(g^{-1}ag)+g^{-1}ag^{2}f(g^{-1}bg)$. After that how can i proceed.@ParthivBasu – math Jul 27 '19 at 13:30
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    I think the difference is in whether $M$ is a left- or right $G$-module. In Silverman, $M$ is a right $G$-module. You've changed the notation from what's in Silverman (if you were looking at Appendix B, as I am): I think it should really be $f(ab) = f(a) b + f(b)$. In many other sources on group cohomology, $M$ is instead a left $G$-module, so the second definition is correct. In Galois cohomology the Galois group naturally acts on the right which is why Silverman considers right modules. – Viktor Vaughn Jan 24 '20 at 08:08

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Claim: $\left( g\cdot f\right) \left( h_{1}h_{2}\right) =h_{1}\left( g\cdot f\right) \left( h_{2}\right) +\left( g\cdot f\right) \left( h_{1}\right) $ for all $h_1 , h_2 \in H$.

Proof. From the other answer you have

$$(g\cdot f)(h)=gf\left(g^{-1}(hg)\right) =gf(g^{-1})+f(hg)=gf(g^{-1})+f(h)+hf(g).$$

Using this equality we get

$$ \begin{align} \ h_{1}\left( g \cdot f\right) \left( h_{2}\right) &=h_{1}[gf\left( g^{-1}\right) +f\left( h_{2}\right) +h_{2}f\left( g\right) ] \\ &= h_{1}gf\left( g^{-1}\right) +h_{1}f\left( h_{2}\right) + h_{1}h_{2}f\left( g\right), \\ \left( g\cdot f\right) \left( h_{1}\right) & =gf\left( g^{-1}\right) +f\left( h_{1}\right) +h_{1}f\left( g\right) \end{align}$$ and $$ \begin{align} \left( g \cdot f\right) \left( h_{1}h_{2}\right) &=gf\left( g^{-1}\right) +f\left( h_{1}h_{2}\right) +h_{1}h_{2}f\left( g\right) \\ &= gf\left( g^{-1}\right) +h_{1}f\left( h_{2}\right) +f\left( h_{1}\right) +h_{1}h_{2}f\left( g\right). \end{align} $$

Note that $h_{1}[ f\left( g\right) +gf\left( g^{-1}\right)] = 0$, so the LHS is indeed equal to the RHS.

Parthiv Basu
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