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Let $M$ and $G$ be groups. We call $M$ a $G$-group (or group with operators) if every $g \in G$ corresponds to an endomorphism of $M$, i.e. we have $$ (mn)^g = (n^g)(m^g). $$ (the application of elements from $G$ in exponential notation). Now let $M$ be abelian, and we write it additively. If $R$ is an associative and on both sides distributive ring with unit $1$, and $M$ is an $R$-group in the above sense, i.e we have

(1) $(m_1 + m_2)r = m_1 r + m_2 r$

for all $r \in R$ and $m_1, m_2 \in M$. And suppose further we have

(2) $m1 = m$

(3) $(mr)r' = m(rr')$

(4) $m(r+r') = mr + mr'$

for $m \in M$, $r,r' \in R$. Then we call $M$ an $R$-right module. I have a question on the following statement:

If $M$ is an abelian $G$-group with $m1 = m$ for all $m \in M$, then $M$ is a $\mathbb Z[G]$-module by $$ m \sum_{g \in G} a_g g := \sum_{g\in G} a_g (mg). $$

I do not understand how $M$ becomes a $\mathbb Z[G]$-module, if we just have $(m_1 + m_2)r = m_1 r + m_2r$ and $m1 = m$. For example why (3) of the above definition holds, I do not even see that for $g, h \in G$ we have $(mg)h = m(gh)$?

These definitions are taken almost verbally from B. Huppert, Endliche Gruppen, unfortunately the relevant pages are not visible from google books. My other recent question about g linear actions was related to thinkings about the above statement. And of course I see that every abelian group could be regarded as a $\mathbb Z$-module (in a unique way, as if $M$ is a $\mathbb Z$-module, then $m1 = m$, hence $m(k+1) = mk + m$ and $m - m = 0 = (1-1)m = m - (-1)m$, hence $-m = (-1)m$, which gives that $mk = m + \ldots + m$ ($k$-times), i.e. its equals the induced $\mathbb Z$-module structure). So abelian groups and $\mathbb Z$-modules are in essence the same. But I do not see if this might be helpful here.

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With the definitions you give, the statement you are asking about can very very easily be false. As you observe, it will be false for any $G$-group structure on $M$ such that $(mg)h\neq m(gh)$ for some $m\in M$ and $g,h\in G$. It is trivial to come up with an example of such a $G$-group; for instance, you could take $G$ to be any nontrivial group, $M=\mathbb{Z}$, and define $m1=m$ and $mg=2m$ for each $m\in M$ and each $g\in G\setminus\{1\}$. Then if $g,h\in G\setminus\{1\}$, $(mg)h=4m$ for all $m$, but $m(gh)$ is either $2m$ or $m$ (depending on whether $gh=1$).

That said, I am 99% certain that the intended definition of "$G$-group" includes the condition that $(mg)h=m(gh)$. While this is not the part of the general definition of a group with a set of operators, when the collection of operators forms a group, you want to add this condition as an additional axiom in the definition (otherwise, why are you bothering to consider the set of operators as a group at all; you're not using its group structure in any way). If Huppert's book didn't state this in its definition, this was an error.

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  • $\begingroup$ Thanks for your answer! But $m1 = m$ was assumed, but not $(mg)h = m(gh)$, so your example is not valid. But what might work is $G = \{1,g\}$ with $mg = 2m$, then $m(gg) = m(1) = m$, but $(mg)g = 4m$. $\endgroup$ – StefanH Jan 24 '16 at 11:01
  • $\begingroup$ Sure, that works. I've edited my example to not use the identity of $G$. $\endgroup$ – Eric Wofsey Jan 24 '16 at 11:05
  • $\begingroup$ Okay, thanks. I thought about the reverse, i.e. requiring $(mg)h = m(gh)$ but dropping $m1 = m$, but could not find any example, I opened a new question for it: math.stackexchange.com/questions/1624791/… $\endgroup$ – StefanH Jan 24 '16 at 11:25

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