I somehow came through this True or False question, but before answering it, I just wanted to ensure some things.
I understand that the axioms for $group-action$ must be met. Basically:
(Compatibility with identity): e∗x=x for all x∈X
(Compatibility with group operation): g1∗(g2∗x)=(g1∗g2)∗x,∀x∈X,g1,g2∈G.
where $*$ is the operation, $X$ the set and $G$ the group.
My question is, for the first axiom, the identity $e$ is obtained from the Group or from the set?
Any help would be appreciated.
It's not a group action.
A group action of $G$ on a set $X$ is a map $G\times X\rightarrow X$ that satisfies a list of conditions, in particular, that $g_1.(g_2.x) = (g_1g_2).x$.
In your situation, take $n = 2$, $x = 1$, then the inverse of $n$ in $\mathbb{Z}$ is $n^{-1} = -2$, but $n^{-1}.(n.x) = (-2)\cdot (2\cdot 1) = -4$, but $(n^{-1}n).x = (-2 + 2)\cdot 1 = 0$.
To answer your question though, a set doesn't come with the data of a distinguished "identity element". Here, you're considering $\mathbb{R}$ as a set, so you're completely forgetting about its group structure as far as the definition of your "action" is concerned.