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Let $M$ be an abelian group and let $G$ be a group acting on $M$ such that $M$ is a $G$-operator group, i.e. we have for $u, v \in M$ and $g,h \in G$

(1) $u\cdot 1_G = u$

(2) $(ug)h = u(gh)$

(3) $(u+v)g = ug + vg$

If $M$ is also a $\mathbb F$-vector space, then we say that $G$ is a linear group action if also we have

(4) $(\lambda u)g = \lambda(ug)$

for $\lambda \in \mathbb F$ and $g \in G$, $u \in M$. In both cases, depending on context, we speak about $G$-modules, this is also mentioned on the linked wikipedia-article. By $G$-module I mean here that only (1), (2) and (3) hold.

Do you know any example where these notions are indeed different, i.e. an example where (1), (2) and (3) holds, but (4) fails?

I tried to deduce (4) somehow from the other, my approach: Let $M$ and $G$ be given such that only (1), (2) and (3) is true. Then we can extend this to an action of the group ring $\mathbb F[G]$ (the set of formal sums, or functions $G \to \mathbb F$) by defining $$ v \cdot \left( \sum_{a_g} a_g g \right) := \sum_{g\in G} a_g (vg) $$ for $v \in M$. Here we have just used the action of $G$ on $M$ and that $V$ is a $\mathbb F$-vector space. Then we could embed $\mathbb F \cdot 1_G \subseteq \mathbb F[G]$ and by (1) the induced action of $\mathbb F$ in $\mathbb F[G]$ and the one already given coincide, as $\lambda v = v\lambda$ and \begin{align*} \lambda v & = \lambda (v \cdot 1_G) & \mbox{by (1)} \\ & = v \cdot (\lambda \cdot 1_G) & \mbox{Definition of extended action.} \\ \end{align*} And taking this further (just a thought, the computations are actually wrong) \begin{align*} (\lambda v) g & = (v(\lambda \cdot 1_G)) g & \mbox{by the above} \\ & = v( (\lambda \cdot 1_G)g ) & \mbox{by (2)} \\ & = v ( \lambda g ) & \mbox{computation in $\mathbb F[G]$}. \end{align*} Okay, the step in the second line was a little bit cheating, as (2) just holds for group elements, and not elements from $\mathbb F[G]$ as applied here. So this might not work and they are indeed different, but I cannot think of any examples?

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Let $M=\mathbb{F}=\mathbb{C}$ and $G$ a cyclic group of order two generated by $g$. Then there is an action of $G$ on $M$ where $g$ acts by complex conjugation, that satisfies (1)-(3), but not (4).

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