I am working on an exercise in group cohomology and need the following result.
Prove that there is a unique group action of $\mathbb{Z}_2$ on $\mathbb{Z}_2$.
I think what's confusing me most is that I believe I found two group actions, the trivial action and the following
$\alpha:\mathbb{Z}_2\times \mathbb{Z}_2\rightarrow \mathbb{Z}_2 \; \; \ni \alpha(0,0) = 0, \; \alpha(0,1) = 1, \; \alpha(1,0) = 1, \; \alpha(1,1) = 0$
Is this action somehow isomorphic to the trivial action? If so, how can I see that?
I think this is about an action of a group on an (abelian) group through group automorphisms. The way the question was worded was potentially confusing but I suspect it came from a context where the intention was clear. Group cohomology is mostly about situations where you have chain complexes of abelian groups with actions of a group by automorphisms (having said that, there are some situations where abelian groups are replaced by groups ore even sets).
A similar situation could occur with the expression a group action on a ring - the intention would almost certainly be that it was an action through ring automorphisms.
But if/when in doubt check!