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Suppose a group $G$ is acting on an abelian group $M$. Then a mapping $\phi: G \rightarrow M$ is called a crossed homomorphism if it satisfies the condition: $\phi(gh)=\phi(g)(g\cdot \phi(h))$ for every $g,h\in G$. My question is, how we will specify the action of $G$ is left or right in the definition of a crossed homomorphism from $G$ to $M$? I found these definitions: If the action is left then we write $\phi(gh)=\phi(g)(g\cdot \phi(h))$. If the action is right, then we write $\phi(gh)=(\phi(g)\cdot h)\phi(h)$. My doubt is, if the action is right, Why not $\phi(gh)=\phi(g)(\phi(h)\cdot g)$ ? Can anyone please clear this concept of left or right?

Kavita
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Normally the easiest way to deal with such problems is to avoid ever considering right group actions. Suppose that we are given a group $G$ and a right $G$-module $M$ with action $\star$

Then we may make $M$ a left $G$-module by defining a left $G$-action via $$g \cdot m = m \star g^{-1}$$

Edit: With the clarifying comments (thankyou @Derek Holt and my apologies to those who enjoy right actions) it seems useful to add futher explanation.

If $\phi : G \to M$ is a cocycle under this induced left action then the induced map $\psi : G \to M$ (remembering that we are first undergoing the automorphism $G \to G : g \mapsto g^{-1}$ ) satisfies the following condition \begin{align*} \psi(gh) = \phi((gh)^{-1}) &= \phi(h^{-1}g^{-1}) \\ &= h^{-1} \cdot \phi(g^{-1}) + \phi(h^{-1}) \\ &= \psi(g)\star h + \psi(h) \end{align*} which is precisely the cocycle condition we are claiming.

Mummy the turkey
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    I agree with that. But in the definitions, the role of g and h get interchanged. Why? – Kavita Jun 28 '20 at 08:46
  • @Kavita hmm yes this wierdly doesn't seem to agree with what I have written (to my detriment I must admit I didn't bother to check). It would help if you could provide a reference. I checked a few of the standard ones and couldn't find anyone who could be bothered with right modules – Mummy the turkey Jun 28 '20 at 22:43
  • The similar question had asked earlier..It might helps to find the reason. https://math.stackexchange.com/q/3305095/738599 – Kavita Jun 29 '20 at 04:54
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    Actually this does answer the question, because $(gh)^{-1} = h^{-1}g^{-1}$, and so the order is reversed when you invert a product of two elements. (But many of us prefer right actions to left actions, so we have no desire at all to avoid considering them!) – Derek Holt Jun 29 '20 at 07:54
  • Thank you both for clearing the concept :) – Kavita Jun 29 '20 at 10:58
  • Is there a literature reference for crossed homomorphism? – Jim Stasheff May 27 '22 at 15:16